Examples of subgroups where it's nontrivial to show closure under multiplication?Can a finitely generated group have infinitely many torsion elements?The torsion subset of a non-abelian group is not, in general, a subgroup.If there exists a subgroup contained in all nontrivial subgroups of $G$, then $G$ is quasicyclic.torsion subgroup of a finitely generated nilpotent group is finite.Is a finitely generated torsion group finite in general?Residually finite group with finitely many conjugacy classes of elements of finite orderPart of simple proof of nontrivial center in p-groupWhy $|G:Z(G)|$ is finite in this question?Is the FC-center of a finitely generated group itself finitely generated?Conjugacy classes generating an infinite group
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Examples of subgroups where it's nontrivial to show closure under multiplication?
Can a finitely generated group have infinitely many torsion elements?The torsion subset of a non-abelian group is not, in general, a subgroup.If there exists a subgroup contained in all nontrivial subgroups of $G$, then $G$ is quasicyclic.torsion subgroup of a finitely generated nilpotent group is finite.Is a finitely generated torsion group finite in general?Residually finite group with finitely many conjugacy classes of elements of finite orderPart of simple proof of nontrivial center in p-groupWhy $|G:Z(G)|$ is finite in this question?Is the FC-center of a finitely generated group itself finitely generated?Conjugacy classes generating an infinite group
$begingroup$
Usually when a subgroup is declared, it is trivial (or at least straightforward to a sophomore) to prove that it is a subgroup under multiplication. For example:
- Homomorphic image and preimage of a subgroup
- Center
- Intersection of two subgroups
- Stabilizer of a point in a group action
- Elements of finite conjugacy class
$HN$, where $Hleq G$ and $Ntrianglelefteq G$
I'm looking for interesting theorems where a subset is claimed to be a subgroup, but it is nontrivial to verify. I am looking for some kind of structure-theoretic subgroup that can be defined for any group (or a large class of groups), rather than specific examples that are hard to show closure.
I can think of only one example.
Let $Delta(G)$ be those elements with finite conjugacy class (easily seen to be a subgroup), and let $Delta^+(G)$ be its torsion subset. Note that $g$ has finite conjugacy class in $G$ if and only if $[G:C(g)]<infty$ where $C(g)$ is the centralizer.
$$Delta^+(G):=gin G : .$$
Theorem: $Delta^+(G)$ is closed under multiplication.
The proof takes 1-2 pages of nontrivial calculations. If $a,b$ are torsion, it's not true that their product $ab$ is torsion --- but amazingly, it is true if $a,b$ have only finitely many conjugates.
Does anyone have any other examples?
Proof of the Theorem, for those interested. You can see it from the following (nontrivial) lemma.
Lemma (Dietzmann). If $[G:Z(G)]<infty$, then $[G,G]$ is finite.
Modulo the (page long) proof of this, let's see why it implies the theorem.
Let $x,yin Delta^+(G)$ so that $x,y$ have finite conjugacy classes and orders. Clearly $xy$ has a finite conjugacy class, so we just have to show that it has finite order.
Let $N$ be the subgroup generated by all the conjugates of $x$ and $y$, so that $N$ is finitely-generated. Then $N/N'$ is an abelian group generated by finitely many torsion elements, hence finite, so $[N:N']<infty$. It is thus enough to show $N'$ is finite, because then $N$ is finite, and since $xyin N$ this completes the proof.
To show $N'$ is finite we use Dietzmann's Lemma: notice $Z(N)=C_N(x)cap C_N(y)$, and these centralizers have finite index in $N$. Therefore $[N:Z(N)]<infty$ and we apply Dietzmann's Lemma.
This was already a somewhat lengthy and interesting argument, and we haven't even proved Dietzmann's Lemma yet!
Edit: a related question is as follows. Name any functions $varphi:Grightarrow H$ that are homomorphisms, but it is nontrivial to show.
abstract-algebra group-theory examples-counterexamples big-list
$endgroup$
|
show 8 more comments
$begingroup$
Usually when a subgroup is declared, it is trivial (or at least straightforward to a sophomore) to prove that it is a subgroup under multiplication. For example:
- Homomorphic image and preimage of a subgroup
- Center
- Intersection of two subgroups
- Stabilizer of a point in a group action
- Elements of finite conjugacy class
$HN$, where $Hleq G$ and $Ntrianglelefteq G$
I'm looking for interesting theorems where a subset is claimed to be a subgroup, but it is nontrivial to verify. I am looking for some kind of structure-theoretic subgroup that can be defined for any group (or a large class of groups), rather than specific examples that are hard to show closure.
I can think of only one example.
Let $Delta(G)$ be those elements with finite conjugacy class (easily seen to be a subgroup), and let $Delta^+(G)$ be its torsion subset. Note that $g$ has finite conjugacy class in $G$ if and only if $[G:C(g)]<infty$ where $C(g)$ is the centralizer.
$$Delta^+(G):=gin G : .$$
Theorem: $Delta^+(G)$ is closed under multiplication.
The proof takes 1-2 pages of nontrivial calculations. If $a,b$ are torsion, it's not true that their product $ab$ is torsion --- but amazingly, it is true if $a,b$ have only finitely many conjugates.
Does anyone have any other examples?
Proof of the Theorem, for those interested. You can see it from the following (nontrivial) lemma.
Lemma (Dietzmann). If $[G:Z(G)]<infty$, then $[G,G]$ is finite.
Modulo the (page long) proof of this, let's see why it implies the theorem.
Let $x,yin Delta^+(G)$ so that $x,y$ have finite conjugacy classes and orders. Clearly $xy$ has a finite conjugacy class, so we just have to show that it has finite order.
Let $N$ be the subgroup generated by all the conjugates of $x$ and $y$, so that $N$ is finitely-generated. Then $N/N'$ is an abelian group generated by finitely many torsion elements, hence finite, so $[N:N']<infty$. It is thus enough to show $N'$ is finite, because then $N$ is finite, and since $xyin N$ this completes the proof.
To show $N'$ is finite we use Dietzmann's Lemma: notice $Z(N)=C_N(x)cap C_N(y)$, and these centralizers have finite index in $N$. Therefore $[N:Z(N)]<infty$ and we apply Dietzmann's Lemma.
This was already a somewhat lengthy and interesting argument, and we haven't even proved Dietzmann's Lemma yet!
Edit: a related question is as follows. Name any functions $varphi:Grightarrow H$ that are homomorphisms, but it is nontrivial to show.
abstract-algebra group-theory examples-counterexamples big-list
$endgroup$
$begingroup$
The set of torsion elements in a nilpotent group is a subgroup. And the set of elements of order some power of a prime $p$, in a nilpotent group, is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:26
$begingroup$
The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:28
1
$begingroup$
I don't remember the precise statement, but there is a theorem that looks like "let the finite group $G$ act transitively on the finite set $X$ such that every element of $G$ has at most one fixed point. Then the set of elements of $G$ that have no fixed point + $e$ is a subgroup of $G$" - if I remember correctly, the proof uses representation theory
$endgroup$
– Max
Apr 27 at 14:31
1
$begingroup$
You don't need any use of structure theorem for f.g. abelian groups (clearly every f.g. abelian group generated by torsion elements is finite).
$endgroup$
– YCor
Apr 27 at 14:37
2
$begingroup$
I don't know if it qualifies, but it is a very difficult and recent theorem (positive solution to Ore's conjecture) that the set of commutators in every finite simple group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:41
|
show 8 more comments
$begingroup$
Usually when a subgroup is declared, it is trivial (or at least straightforward to a sophomore) to prove that it is a subgroup under multiplication. For example:
- Homomorphic image and preimage of a subgroup
- Center
- Intersection of two subgroups
- Stabilizer of a point in a group action
- Elements of finite conjugacy class
$HN$, where $Hleq G$ and $Ntrianglelefteq G$
I'm looking for interesting theorems where a subset is claimed to be a subgroup, but it is nontrivial to verify. I am looking for some kind of structure-theoretic subgroup that can be defined for any group (or a large class of groups), rather than specific examples that are hard to show closure.
I can think of only one example.
Let $Delta(G)$ be those elements with finite conjugacy class (easily seen to be a subgroup), and let $Delta^+(G)$ be its torsion subset. Note that $g$ has finite conjugacy class in $G$ if and only if $[G:C(g)]<infty$ where $C(g)$ is the centralizer.
$$Delta^+(G):=gin G : .$$
Theorem: $Delta^+(G)$ is closed under multiplication.
The proof takes 1-2 pages of nontrivial calculations. If $a,b$ are torsion, it's not true that their product $ab$ is torsion --- but amazingly, it is true if $a,b$ have only finitely many conjugates.
Does anyone have any other examples?
Proof of the Theorem, for those interested. You can see it from the following (nontrivial) lemma.
Lemma (Dietzmann). If $[G:Z(G)]<infty$, then $[G,G]$ is finite.
Modulo the (page long) proof of this, let's see why it implies the theorem.
Let $x,yin Delta^+(G)$ so that $x,y$ have finite conjugacy classes and orders. Clearly $xy$ has a finite conjugacy class, so we just have to show that it has finite order.
Let $N$ be the subgroup generated by all the conjugates of $x$ and $y$, so that $N$ is finitely-generated. Then $N/N'$ is an abelian group generated by finitely many torsion elements, hence finite, so $[N:N']<infty$. It is thus enough to show $N'$ is finite, because then $N$ is finite, and since $xyin N$ this completes the proof.
To show $N'$ is finite we use Dietzmann's Lemma: notice $Z(N)=C_N(x)cap C_N(y)$, and these centralizers have finite index in $N$. Therefore $[N:Z(N)]<infty$ and we apply Dietzmann's Lemma.
This was already a somewhat lengthy and interesting argument, and we haven't even proved Dietzmann's Lemma yet!
Edit: a related question is as follows. Name any functions $varphi:Grightarrow H$ that are homomorphisms, but it is nontrivial to show.
abstract-algebra group-theory examples-counterexamples big-list
$endgroup$
Usually when a subgroup is declared, it is trivial (or at least straightforward to a sophomore) to prove that it is a subgroup under multiplication. For example:
- Homomorphic image and preimage of a subgroup
- Center
- Intersection of two subgroups
- Stabilizer of a point in a group action
- Elements of finite conjugacy class
$HN$, where $Hleq G$ and $Ntrianglelefteq G$
I'm looking for interesting theorems where a subset is claimed to be a subgroup, but it is nontrivial to verify. I am looking for some kind of structure-theoretic subgroup that can be defined for any group (or a large class of groups), rather than specific examples that are hard to show closure.
I can think of only one example.
Let $Delta(G)$ be those elements with finite conjugacy class (easily seen to be a subgroup), and let $Delta^+(G)$ be its torsion subset. Note that $g$ has finite conjugacy class in $G$ if and only if $[G:C(g)]<infty$ where $C(g)$ is the centralizer.
$$Delta^+(G):=gin G : .$$
Theorem: $Delta^+(G)$ is closed under multiplication.
The proof takes 1-2 pages of nontrivial calculations. If $a,b$ are torsion, it's not true that their product $ab$ is torsion --- but amazingly, it is true if $a,b$ have only finitely many conjugates.
Does anyone have any other examples?
Proof of the Theorem, for those interested. You can see it from the following (nontrivial) lemma.
Lemma (Dietzmann). If $[G:Z(G)]<infty$, then $[G,G]$ is finite.
Modulo the (page long) proof of this, let's see why it implies the theorem.
Let $x,yin Delta^+(G)$ so that $x,y$ have finite conjugacy classes and orders. Clearly $xy$ has a finite conjugacy class, so we just have to show that it has finite order.
Let $N$ be the subgroup generated by all the conjugates of $x$ and $y$, so that $N$ is finitely-generated. Then $N/N'$ is an abelian group generated by finitely many torsion elements, hence finite, so $[N:N']<infty$. It is thus enough to show $N'$ is finite, because then $N$ is finite, and since $xyin N$ this completes the proof.
To show $N'$ is finite we use Dietzmann's Lemma: notice $Z(N)=C_N(x)cap C_N(y)$, and these centralizers have finite index in $N$. Therefore $[N:Z(N)]<infty$ and we apply Dietzmann's Lemma.
This was already a somewhat lengthy and interesting argument, and we haven't even proved Dietzmann's Lemma yet!
Edit: a related question is as follows. Name any functions $varphi:Grightarrow H$ that are homomorphisms, but it is nontrivial to show.
abstract-algebra group-theory examples-counterexamples big-list
abstract-algebra group-theory examples-counterexamples big-list
edited 2 days ago
Martin Sleziak
45.2k11123278
45.2k11123278
asked Apr 27 at 14:07
EhsaanEhsaan
1,171515
1,171515
$begingroup$
The set of torsion elements in a nilpotent group is a subgroup. And the set of elements of order some power of a prime $p$, in a nilpotent group, is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:26
$begingroup$
The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:28
1
$begingroup$
I don't remember the precise statement, but there is a theorem that looks like "let the finite group $G$ act transitively on the finite set $X$ such that every element of $G$ has at most one fixed point. Then the set of elements of $G$ that have no fixed point + $e$ is a subgroup of $G$" - if I remember correctly, the proof uses representation theory
$endgroup$
– Max
Apr 27 at 14:31
1
$begingroup$
You don't need any use of structure theorem for f.g. abelian groups (clearly every f.g. abelian group generated by torsion elements is finite).
$endgroup$
– YCor
Apr 27 at 14:37
2
$begingroup$
I don't know if it qualifies, but it is a very difficult and recent theorem (positive solution to Ore's conjecture) that the set of commutators in every finite simple group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:41
|
show 8 more comments
$begingroup$
The set of torsion elements in a nilpotent group is a subgroup. And the set of elements of order some power of a prime $p$, in a nilpotent group, is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:26
$begingroup$
The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:28
1
$begingroup$
I don't remember the precise statement, but there is a theorem that looks like "let the finite group $G$ act transitively on the finite set $X$ such that every element of $G$ has at most one fixed point. Then the set of elements of $G$ that have no fixed point + $e$ is a subgroup of $G$" - if I remember correctly, the proof uses representation theory
$endgroup$
– Max
Apr 27 at 14:31
1
$begingroup$
You don't need any use of structure theorem for f.g. abelian groups (clearly every f.g. abelian group generated by torsion elements is finite).
$endgroup$
– YCor
Apr 27 at 14:37
2
$begingroup$
I don't know if it qualifies, but it is a very difficult and recent theorem (positive solution to Ore's conjecture) that the set of commutators in every finite simple group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:41
$begingroup$
The set of torsion elements in a nilpotent group is a subgroup. And the set of elements of order some power of a prime $p$, in a nilpotent group, is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:26
$begingroup$
The set of torsion elements in a nilpotent group is a subgroup. And the set of elements of order some power of a prime $p$, in a nilpotent group, is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:26
$begingroup$
The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:28
$begingroup$
The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:28
1
1
$begingroup$
I don't remember the precise statement, but there is a theorem that looks like "let the finite group $G$ act transitively on the finite set $X$ such that every element of $G$ has at most one fixed point. Then the set of elements of $G$ that have no fixed point + $e$ is a subgroup of $G$" - if I remember correctly, the proof uses representation theory
$endgroup$
– Max
Apr 27 at 14:31
$begingroup$
I don't remember the precise statement, but there is a theorem that looks like "let the finite group $G$ act transitively on the finite set $X$ such that every element of $G$ has at most one fixed point. Then the set of elements of $G$ that have no fixed point + $e$ is a subgroup of $G$" - if I remember correctly, the proof uses representation theory
$endgroup$
– Max
Apr 27 at 14:31
1
1
$begingroup$
You don't need any use of structure theorem for f.g. abelian groups (clearly every f.g. abelian group generated by torsion elements is finite).
$endgroup$
– YCor
Apr 27 at 14:37
$begingroup$
You don't need any use of structure theorem for f.g. abelian groups (clearly every f.g. abelian group generated by torsion elements is finite).
$endgroup$
– YCor
Apr 27 at 14:37
2
2
$begingroup$
I don't know if it qualifies, but it is a very difficult and recent theorem (positive solution to Ore's conjecture) that the set of commutators in every finite simple group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:41
$begingroup$
I don't know if it qualifies, but it is a very difficult and recent theorem (positive solution to Ore's conjecture) that the set of commutators in every finite simple group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:41
|
show 8 more comments
3 Answers
3
active
oldest
votes
$begingroup$
A particularly nice example is the following : suppose the finite group $G$ acts on the finite set $X$ in such a way that every nontrivial element of $G$ has at most one fixed point. Let $S$ be the set of elements of $G$ that have no fixed points. Then $H=Scup 1$ is a subgroup of $G$.
I believe the only known proofs are representation-theoretic (or at least that was the case at first).
$endgroup$
5
$begingroup$
Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
$endgroup$
– Derek Holt
Apr 27 at 16:56
$begingroup$
For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
$endgroup$
– j.p.
2 days ago
add a comment |
$begingroup$
Consider the symmetric group $S_n$ on $ngeq 2$ letters. The alternating group $A_n$ is the subgroup of $S_n$ given by all even permutations of $S_n$.
One proof uses the signum or sign function $s:S_nrightarrowpm 1$ which assigns to a permutation $pi$, $+1$ if $pi$ is even, and $-1$ if $pi$ is odd.
It can be shown that the sign function is a homomorphism, i.e., $s(pisigma) = s(pi)cdot s(sigma)$.
It follows that the product of two even permutations is even and so the multiplication in $A_n$ is well-defined.
$endgroup$
$begingroup$
Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
$endgroup$
– Ehsaan
Apr 27 at 14:31
$begingroup$
When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
$endgroup$
– Ehsaan
Apr 27 at 14:36
$begingroup$
@Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
$endgroup$
– darij grinberg
Apr 27 at 14:53
$begingroup$
@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
$endgroup$
– Ehsaan
Apr 27 at 14:57
2
$begingroup$
An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
$endgroup$
– Ehsaan
Apr 27 at 16:24
|
show 2 more comments
$begingroup$
Let $G=GL(4,k)$ (the group of all $4times4$ invertible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$beginbmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\0&0&a_33&a_34\0&0&a_43&a_44endbmatrix.$$
$endgroup$
4
$begingroup$
I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
$endgroup$
– yamete kudasai
Apr 27 at 14:38
$begingroup$
Indeed it does, but many Linear Algebra students don't know that.
$endgroup$
– José Carlos Santos
Apr 27 at 16:05
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A particularly nice example is the following : suppose the finite group $G$ acts on the finite set $X$ in such a way that every nontrivial element of $G$ has at most one fixed point. Let $S$ be the set of elements of $G$ that have no fixed points. Then $H=Scup 1$ is a subgroup of $G$.
I believe the only known proofs are representation-theoretic (or at least that was the case at first).
$endgroup$
5
$begingroup$
Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
$endgroup$
– Derek Holt
Apr 27 at 16:56
$begingroup$
For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
$endgroup$
– j.p.
2 days ago
add a comment |
$begingroup$
A particularly nice example is the following : suppose the finite group $G$ acts on the finite set $X$ in such a way that every nontrivial element of $G$ has at most one fixed point. Let $S$ be the set of elements of $G$ that have no fixed points. Then $H=Scup 1$ is a subgroup of $G$.
I believe the only known proofs are representation-theoretic (or at least that was the case at first).
$endgroup$
5
$begingroup$
Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
$endgroup$
– Derek Holt
Apr 27 at 16:56
$begingroup$
For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
$endgroup$
– j.p.
2 days ago
add a comment |
$begingroup$
A particularly nice example is the following : suppose the finite group $G$ acts on the finite set $X$ in such a way that every nontrivial element of $G$ has at most one fixed point. Let $S$ be the set of elements of $G$ that have no fixed points. Then $H=Scup 1$ is a subgroup of $G$.
I believe the only known proofs are representation-theoretic (or at least that was the case at first).
$endgroup$
A particularly nice example is the following : suppose the finite group $G$ acts on the finite set $X$ in such a way that every nontrivial element of $G$ has at most one fixed point. Let $S$ be the set of elements of $G$ that have no fixed points. Then $H=Scup 1$ is a subgroup of $G$.
I believe the only known proofs are representation-theoretic (or at least that was the case at first).
answered Apr 27 at 15:58
MaxMax
17.2k11144
17.2k11144
5
$begingroup$
Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
$endgroup$
– Derek Holt
Apr 27 at 16:56
$begingroup$
For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
$endgroup$
– j.p.
2 days ago
add a comment |
5
$begingroup$
Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
$endgroup$
– Derek Holt
Apr 27 at 16:56
$begingroup$
For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
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– j.p.
2 days ago
5
5
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Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
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– Derek Holt
Apr 27 at 16:56
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Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
$endgroup$
– Derek Holt
Apr 27 at 16:56
$begingroup$
For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
$endgroup$
– j.p.
2 days ago
$begingroup$
For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
$endgroup$
– j.p.
2 days ago
add a comment |
$begingroup$
Consider the symmetric group $S_n$ on $ngeq 2$ letters. The alternating group $A_n$ is the subgroup of $S_n$ given by all even permutations of $S_n$.
One proof uses the signum or sign function $s:S_nrightarrowpm 1$ which assigns to a permutation $pi$, $+1$ if $pi$ is even, and $-1$ if $pi$ is odd.
It can be shown that the sign function is a homomorphism, i.e., $s(pisigma) = s(pi)cdot s(sigma)$.
It follows that the product of two even permutations is even and so the multiplication in $A_n$ is well-defined.
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$begingroup$
Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
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– Ehsaan
Apr 27 at 14:31
$begingroup$
When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
$endgroup$
– Ehsaan
Apr 27 at 14:36
$begingroup$
@Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
$endgroup$
– darij grinberg
Apr 27 at 14:53
$begingroup$
@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
$endgroup$
– Ehsaan
Apr 27 at 14:57
2
$begingroup$
An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
$endgroup$
– Ehsaan
Apr 27 at 16:24
|
show 2 more comments
$begingroup$
Consider the symmetric group $S_n$ on $ngeq 2$ letters. The alternating group $A_n$ is the subgroup of $S_n$ given by all even permutations of $S_n$.
One proof uses the signum or sign function $s:S_nrightarrowpm 1$ which assigns to a permutation $pi$, $+1$ if $pi$ is even, and $-1$ if $pi$ is odd.
It can be shown that the sign function is a homomorphism, i.e., $s(pisigma) = s(pi)cdot s(sigma)$.
It follows that the product of two even permutations is even and so the multiplication in $A_n$ is well-defined.
$endgroup$
$begingroup$
Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
$endgroup$
– Ehsaan
Apr 27 at 14:31
$begingroup$
When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
$endgroup$
– Ehsaan
Apr 27 at 14:36
$begingroup$
@Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
$endgroup$
– darij grinberg
Apr 27 at 14:53
$begingroup$
@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
$endgroup$
– Ehsaan
Apr 27 at 14:57
2
$begingroup$
An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
$endgroup$
– Ehsaan
Apr 27 at 16:24
|
show 2 more comments
$begingroup$
Consider the symmetric group $S_n$ on $ngeq 2$ letters. The alternating group $A_n$ is the subgroup of $S_n$ given by all even permutations of $S_n$.
One proof uses the signum or sign function $s:S_nrightarrowpm 1$ which assigns to a permutation $pi$, $+1$ if $pi$ is even, and $-1$ if $pi$ is odd.
It can be shown that the sign function is a homomorphism, i.e., $s(pisigma) = s(pi)cdot s(sigma)$.
It follows that the product of two even permutations is even and so the multiplication in $A_n$ is well-defined.
$endgroup$
Consider the symmetric group $S_n$ on $ngeq 2$ letters. The alternating group $A_n$ is the subgroup of $S_n$ given by all even permutations of $S_n$.
One proof uses the signum or sign function $s:S_nrightarrowpm 1$ which assigns to a permutation $pi$, $+1$ if $pi$ is even, and $-1$ if $pi$ is odd.
It can be shown that the sign function is a homomorphism, i.e., $s(pisigma) = s(pi)cdot s(sigma)$.
It follows that the product of two even permutations is even and so the multiplication in $A_n$ is well-defined.
answered Apr 27 at 14:24
WuestenfuxWuestenfux
5,8991513
5,8991513
$begingroup$
Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
$endgroup$
– Ehsaan
Apr 27 at 14:31
$begingroup$
When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
$endgroup$
– Ehsaan
Apr 27 at 14:36
$begingroup$
@Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
$endgroup$
– darij grinberg
Apr 27 at 14:53
$begingroup$
@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
$endgroup$
– Ehsaan
Apr 27 at 14:57
2
$begingroup$
An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
$endgroup$
– Ehsaan
Apr 27 at 16:24
|
show 2 more comments
$begingroup$
Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
$endgroup$
– Ehsaan
Apr 27 at 14:31
$begingroup$
When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
$endgroup$
– Ehsaan
Apr 27 at 14:36
$begingroup$
@Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
$endgroup$
– darij grinberg
Apr 27 at 14:53
$begingroup$
@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
$endgroup$
– Ehsaan
Apr 27 at 14:57
2
$begingroup$
An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
$endgroup$
– Ehsaan
Apr 27 at 16:24
$begingroup$
Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
$endgroup$
– Ehsaan
Apr 27 at 14:31
$begingroup$
Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
$endgroup$
– Ehsaan
Apr 27 at 14:31
$begingroup$
When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
$endgroup$
– Ehsaan
Apr 27 at 14:36
$begingroup$
When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
$endgroup$
– Ehsaan
Apr 27 at 14:36
$begingroup$
@Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
$endgroup$
– darij grinberg
Apr 27 at 14:53
$begingroup$
@Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
$endgroup$
– darij grinberg
Apr 27 at 14:53
$begingroup$
@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
$endgroup$
– Ehsaan
Apr 27 at 14:57
$begingroup$
@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
$endgroup$
– Ehsaan
Apr 27 at 14:57
2
2
$begingroup$
An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
$endgroup$
– Ehsaan
Apr 27 at 16:24
$begingroup$
An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
$endgroup$
– Ehsaan
Apr 27 at 16:24
|
show 2 more comments
$begingroup$
Let $G=GL(4,k)$ (the group of all $4times4$ invertible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$beginbmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\0&0&a_33&a_34\0&0&a_43&a_44endbmatrix.$$
$endgroup$
4
$begingroup$
I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
$endgroup$
– yamete kudasai
Apr 27 at 14:38
$begingroup$
Indeed it does, but many Linear Algebra students don't know that.
$endgroup$
– José Carlos Santos
Apr 27 at 16:05
add a comment |
$begingroup$
Let $G=GL(4,k)$ (the group of all $4times4$ invertible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$beginbmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\0&0&a_33&a_34\0&0&a_43&a_44endbmatrix.$$
$endgroup$
4
$begingroup$
I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
$endgroup$
– yamete kudasai
Apr 27 at 14:38
$begingroup$
Indeed it does, but many Linear Algebra students don't know that.
$endgroup$
– José Carlos Santos
Apr 27 at 16:05
add a comment |
$begingroup$
Let $G=GL(4,k)$ (the group of all $4times4$ invertible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$beginbmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\0&0&a_33&a_34\0&0&a_43&a_44endbmatrix.$$
$endgroup$
Let $G=GL(4,k)$ (the group of all $4times4$ invertible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$beginbmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\0&0&a_33&a_34\0&0&a_43&a_44endbmatrix.$$
edited 2 days ago
answered Apr 27 at 14:15
José Carlos SantosJosé Carlos Santos
179k24139254
179k24139254
4
$begingroup$
I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
$endgroup$
– yamete kudasai
Apr 27 at 14:38
$begingroup$
Indeed it does, but many Linear Algebra students don't know that.
$endgroup$
– José Carlos Santos
Apr 27 at 16:05
add a comment |
4
$begingroup$
I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
$endgroup$
– yamete kudasai
Apr 27 at 14:38
$begingroup$
Indeed it does, but many Linear Algebra students don't know that.
$endgroup$
– José Carlos Santos
Apr 27 at 16:05
4
4
$begingroup$
I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
$endgroup$
– yamete kudasai
Apr 27 at 14:38
$begingroup$
I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
$endgroup$
– yamete kudasai
Apr 27 at 14:38
$begingroup$
Indeed it does, but many Linear Algebra students don't know that.
$endgroup$
– José Carlos Santos
Apr 27 at 16:05
$begingroup$
Indeed it does, but many Linear Algebra students don't know that.
$endgroup$
– José Carlos Santos
Apr 27 at 16:05
add a comment |
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$begingroup$
The set of torsion elements in a nilpotent group is a subgroup. And the set of elements of order some power of a prime $p$, in a nilpotent group, is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:26
$begingroup$
The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:28
1
$begingroup$
I don't remember the precise statement, but there is a theorem that looks like "let the finite group $G$ act transitively on the finite set $X$ such that every element of $G$ has at most one fixed point. Then the set of elements of $G$ that have no fixed point + $e$ is a subgroup of $G$" - if I remember correctly, the proof uses representation theory
$endgroup$
– Max
Apr 27 at 14:31
1
$begingroup$
You don't need any use of structure theorem for f.g. abelian groups (clearly every f.g. abelian group generated by torsion elements is finite).
$endgroup$
– YCor
Apr 27 at 14:37
2
$begingroup$
I don't know if it qualifies, but it is a very difficult and recent theorem (positive solution to Ore's conjecture) that the set of commutators in every finite simple group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:41