Doubt regarding “Elementary approach to proving that a group of order 9 is Abelian”Elementary approach to proving that a group of order 9 is AbelianGroup of order $p^2$ is abelian.showing that a group of order 45 is abelianElementary approach to proving that a group of order 9 is AbelianGroup of order 175 is AbelianNon-isomorphic non-abelian groups of order 243?$G$ is a group of odd order, show that $a^2=b^2 Rightarrow a=b$Non-abelian group of order $6$To Show that a group is abeliandoubt in proving that an abelian group of order 35 is cyclic.The cases in proving that a group of order 90 is not simple
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Doubt regarding “Elementary approach to proving that a group of order 9 is Abelian”
Elementary approach to proving that a group of order 9 is AbelianGroup of order $p^2$ is abelian.showing that a group of order 45 is abelianElementary approach to proving that a group of order 9 is AbelianGroup of order 175 is AbelianNon-isomorphic non-abelian groups of order 243?$G$ is a group of odd order, show that $a^2=b^2 Rightarrow a=b$Non-abelian group of order $6$To Show that a group is abeliandoubt in proving that an abelian group of order 35 is cyclic.The cases in proving that a group of order 90 is not simple
$begingroup$
I am trying to understand the solution of this problem . I am unable to understand why :
If $yx=x^2y$, then $yxy^-1=x^2$. This means that $y^3xy^-3=x^8 $
It seems like I am missing something silly , but why is this true ?
abstract-algebra group-theory finite-groups
asked May 18 at 15:34
JohnJohn
395113
$endgroup$
|
show 2 more comments
$begingroup$
I am trying to understand the solution of this problem . I am unable to understand why :
If $yx=x^2y$, then $yxy^-1=x^2$. This means that $y^3xy^-3=x^8 $
It seems like I am missing something silly , but why is this true ?
abstract-algebra group-theory finite-groups
asked May 18 at 15:34
JohnJohn
395113
$endgroup$
1
$begingroup$
@Alvin Lepik No, this is not true. We do not suppose that the group is abelian. This is what we need to prove.
$endgroup$
– Mark
May 18 at 15:38
$begingroup$
$(ab)^c = (ab)(ab) cdots (ab) $ Then , why is it so , @Alvin Lepik
$endgroup$
– John
May 18 at 15:39
1
$begingroup$
My comment was wrong. I deleted it.
$endgroup$
– Mark
May 18 at 15:41
1
$begingroup$
Ditto what Mark said.
$endgroup$
– Alvin Lepik
May 18 at 15:47
1
$begingroup$
@Shaun , IThanks for reminding ! I wanted to accept it immediately but the website told me to wait 4 minutes , after which I forgot to mark it as the answer Lol :P
$endgroup$
– John
May 18 at 15:58
|
show 2 more comments
$begingroup$
I am trying to understand the solution of this problem . I am unable to understand why :
If $yx=x^2y$, then $yxy^-1=x^2$. This means that $y^3xy^-3=x^8 $
It seems like I am missing something silly , but why is this true ?
abstract-algebra group-theory finite-groups
asked May 18 at 15:34
JohnJohn
395113
$endgroup$
I am trying to understand the solution of this problem . I am unable to understand why :
If $yx=x^2y$, then $yxy^-1=x^2$. This means that $y^3xy^-3=x^8 $
It seems like I am missing something silly , but why is this true ?
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
asked May 18 at 15:34
JohnJohn
395113
asked May 18 at 15:34
JohnJohn
395113
asked May 18 at 15:34
JohnJohn
395113
asked May 18 at 15:34
JohnJohn
395113
asked May 18 at 15:34
JohnJohn
395113
395113
1
$begingroup$
@Alvin Lepik No, this is not true. We do not suppose that the group is abelian. This is what we need to prove.
$endgroup$
– Mark
May 18 at 15:38
$begingroup$
$(ab)^c = (ab)(ab) cdots (ab) $ Then , why is it so , @Alvin Lepik
$endgroup$
– John
May 18 at 15:39
1
$begingroup$
My comment was wrong. I deleted it.
$endgroup$
– Mark
May 18 at 15:41
1
$begingroup$
Ditto what Mark said.
$endgroup$
– Alvin Lepik
May 18 at 15:47
1
$begingroup$
@Shaun , IThanks for reminding ! I wanted to accept it immediately but the website told me to wait 4 minutes , after which I forgot to mark it as the answer Lol :P
$endgroup$
– John
May 18 at 15:58
|
show 2 more comments
1
$begingroup$
@Alvin Lepik No, this is not true. We do not suppose that the group is abelian. This is what we need to prove.
$endgroup$
– Mark
May 18 at 15:38
$begingroup$
$(ab)^c = (ab)(ab) cdots (ab) $ Then , why is it so , @Alvin Lepik
$endgroup$
– John
May 18 at 15:39
1
$begingroup$
My comment was wrong. I deleted it.
$endgroup$
– Mark
May 18 at 15:41
1
$begingroup$
Ditto what Mark said.
$endgroup$
– Alvin Lepik
May 18 at 15:47
1
$begingroup$
@Shaun , IThanks for reminding ! I wanted to accept it immediately but the website told me to wait 4 minutes , after which I forgot to mark it as the answer Lol :P
$endgroup$
– John
May 18 at 15:58
1
1
$begingroup$
@Alvin Lepik No, this is not true. We do not suppose that the group is abelian. This is what we need to prove.
$endgroup$
– Mark
May 18 at 15:38
$begingroup$
@Alvin Lepik No, this is not true. We do not suppose that the group is abelian. This is what we need to prove.
$endgroup$
– Mark
May 18 at 15:38
$begingroup$
$(ab)^c = (ab)(ab) cdots (ab) $ Then , why is it so , @Alvin Lepik
$endgroup$
– John
May 18 at 15:39
$begingroup$
$(ab)^c = (ab)(ab) cdots (ab) $ Then , why is it so , @Alvin Lepik
$endgroup$
– John
May 18 at 15:39
1
1
$begingroup$
My comment was wrong. I deleted it.
$endgroup$
– Mark
May 18 at 15:41
$begingroup$
My comment was wrong. I deleted it.
$endgroup$
– Mark
May 18 at 15:41
1
1
$begingroup$
Ditto what Mark said.
$endgroup$
– Alvin Lepik
May 18 at 15:47
$begingroup$
Ditto what Mark said.
$endgroup$
– Alvin Lepik
May 18 at 15:47
1
1
$begingroup$
@Shaun , IThanks for reminding ! I wanted to accept it immediately but the website told me to wait 4 minutes , after which I forgot to mark it as the answer Lol :P
$endgroup$
– John
May 18 at 15:58
$begingroup$
@Shaun , IThanks for reminding ! I wanted to accept it immediately but the website told me to wait 4 minutes , after which I forgot to mark it as the answer Lol :P
$endgroup$
– John
May 18 at 15:58
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$$
beginalign
y^3xy^-3&=y^2(yxy^-1)y^-2\
&=y^2x^2y^-2\
&=y(yx^2y^-1)y^-1\
&=y(yxy^-1)^2y^-1\
&=y(x^2)^2y^-1\
&=yx^4y^-1\
&=(yxy^-1)^4\
&=(x^2)^4\
&=x^8
endalign
$$
edited May 18 at 15:54
Shaun
11.2k123688
answered May 18 at 15:44
MarkMark
12.9k1825
$endgroup$
$begingroup$
+1 So this did involve a bit of manipulation !
$endgroup$
– John
May 18 at 15:46
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$
beginalign
y^3xy^-3&=y^2(yxy^-1)y^-2\
&=y^2x^2y^-2\
&=y(yx^2y^-1)y^-1\
&=y(yxy^-1)^2y^-1\
&=y(x^2)^2y^-1\
&=yx^4y^-1\
&=(yxy^-1)^4\
&=(x^2)^4\
&=x^8
endalign
$$
edited May 18 at 15:54
Shaun
11.2k123688
answered May 18 at 15:44
MarkMark
12.9k1825
$endgroup$
$begingroup$
+1 So this did involve a bit of manipulation !
$endgroup$
– John
May 18 at 15:46
add a comment |
$begingroup$
$$
beginalign
y^3xy^-3&=y^2(yxy^-1)y^-2\
&=y^2x^2y^-2\
&=y(yx^2y^-1)y^-1\
&=y(yxy^-1)^2y^-1\
&=y(x^2)^2y^-1\
&=yx^4y^-1\
&=(yxy^-1)^4\
&=(x^2)^4\
&=x^8
endalign
$$
edited May 18 at 15:54
Shaun
11.2k123688
answered May 18 at 15:44
MarkMark
12.9k1825
$endgroup$
$begingroup$
+1 So this did involve a bit of manipulation !
$endgroup$
– John
May 18 at 15:46
add a comment |
$begingroup$
$$
beginalign
y^3xy^-3&=y^2(yxy^-1)y^-2\
&=y^2x^2y^-2\
&=y(yx^2y^-1)y^-1\
&=y(yxy^-1)^2y^-1\
&=y(x^2)^2y^-1\
&=yx^4y^-1\
&=(yxy^-1)^4\
&=(x^2)^4\
&=x^8
endalign
$$
edited May 18 at 15:54
Shaun
11.2k123688
answered May 18 at 15:44
MarkMark
12.9k1825
$endgroup$
$$
beginalign
y^3xy^-3&=y^2(yxy^-1)y^-2\
&=y^2x^2y^-2\
&=y(yx^2y^-1)y^-1\
&=y(yxy^-1)^2y^-1\
&=y(x^2)^2y^-1\
&=yx^4y^-1\
&=(yxy^-1)^4\
&=(x^2)^4\
&=x^8
endalign
$$
edited May 18 at 15:54
Shaun
11.2k123688
answered May 18 at 15:44
MarkMark
12.9k1825
edited May 18 at 15:54
Shaun
11.2k123688
edited May 18 at 15:54
Shaun
11.2k123688
edited May 18 at 15:54
Shaun
11.2k123688
11.2k123688
answered May 18 at 15:44
MarkMark
12.9k1825
answered May 18 at 15:44
MarkMark
12.9k1825
answered May 18 at 15:44
MarkMark
12.9k1825
12.9k1825
$begingroup$
+1 So this did involve a bit of manipulation !
$endgroup$
– John
May 18 at 15:46
add a comment |
$begingroup$
+1 So this did involve a bit of manipulation !
$endgroup$
– John
May 18 at 15:46
$begingroup$
+1 So this did involve a bit of manipulation !
$endgroup$
– John
May 18 at 15:46
$begingroup$
+1 So this did involve a bit of manipulation !
$endgroup$
– John
May 18 at 15:46
add a comment |
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1
$begingroup$
@Alvin Lepik No, this is not true. We do not suppose that the group is abelian. This is what we need to prove.
$endgroup$
– Mark
May 18 at 15:38
$begingroup$
$(ab)^c = (ab)(ab) cdots (ab) $ Then , why is it so , @Alvin Lepik
$endgroup$
– John
May 18 at 15:39
1
$begingroup$
My comment was wrong. I deleted it.
$endgroup$
– Mark
May 18 at 15:41
1
$begingroup$
Ditto what Mark said.
$endgroup$
– Alvin Lepik
May 18 at 15:47
1
$begingroup$
@Shaun , IThanks for reminding ! I wanted to accept it immediately but the website told me to wait 4 minutes , after which I forgot to mark it as the answer Lol :P
$endgroup$
– John
May 18 at 15:58