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If every nonidentity element in a group is of order $2$, the group is abelian [duplicate]

Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.Prove that a group G such that every element $g in G$ satisfies the equality $g^2 = 1_G$ is abelianProve that a finite abelian group is simple if and only if its order is prime.Every infinite abelian group has at least one element of infinite order?Abelian group and element orderNoncyclic Abelian Group of order 51Prove that it is impossible that every non-identity element of $G$ has an order of $2$.A non-cyclic group $G$ such that every non-identity element $xin G$ has the order $p$ (p prime)How many subgroups of order $3$ does a non-abelian group of order $39$ have?Non-identity element in a group has infinite orderShowing there is a unique group table for $1, a,b,c$ such that there is no element of order $4$.

7

$begingroup$

This question already has an answer here:

• 
  Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.
  
  13 answers
  
  

Let $G$ be a group. Show that if every non-identity element in $G$ has order $2$ then $G$ is abelian.

Proof:

Let $a,b $ be non-identity elements in $G$. Since $|a|=|b|=2$ , that means $ab=babaab$ $=$ $ba$.

Is the proof correct? How can I improve it?

abstract-algebra group-theory proof-verification abelian-groups

share|cite|improve this question

edited 2 days ago

TheSimpliFire

13.9k62765

asked May 4 at 21:03

topologicalmagiciantopologicalmagician

3299

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yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The proof is correct as written.
  $endgroup$
  – rubikscube09
  May 4 at 21:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I must be tired by I don't see how to justfify the first equality.
  $endgroup$
  – elidiot
  May 4 at 21:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @elidiot (baba)ab=(e)ab=ab. bab(aa)b=babb=ba
  $endgroup$
  – topologicalmagician
  May 4 at 21:09
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It uses $|ba|=2$ on the left.
  $endgroup$
  – Berci
  May 4 at 21:10
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @rschwieb thanks, i'll keep that in mind next time.
  $endgroup$
  – topologicalmagician
  12 hours ago
  

  
  
  
  

 | 
show 9 more comments

7

$begingroup$

This question already has an answer here:

• 
  Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.
  
  13 answers
  
  

Let $G$ be a group. Show that if every non-identity element in $G$ has order $2$ then $G$ is abelian.

Proof:

Let $a,b $ be non-identity elements in $G$. Since $|a|=|b|=2$ , that means $ab=babaab$ $=$ $ba$.

Is the proof correct? How can I improve it?

abstract-algebra group-theory proof-verification abelian-groups

share|cite|improve this question

edited 2 days ago

TheSimpliFire

13.9k62765

asked May 4 at 21:03

topologicalmagiciantopologicalmagician

3299

$endgroup$

marked as duplicate by rschwieb abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.

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yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The proof is correct as written.
  $endgroup$
  – rubikscube09
  May 4 at 21:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I must be tired by I don't see how to justfify the first equality.
  $endgroup$
  – elidiot
  May 4 at 21:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @elidiot (baba)ab=(e)ab=ab. bab(aa)b=babb=ba
  $endgroup$
  – topologicalmagician
  May 4 at 21:09
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It uses $|ba|=2$ on the left.
  $endgroup$
  – Berci
  May 4 at 21:10
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @rschwieb thanks, i'll keep that in mind next time.
  $endgroup$
  – topologicalmagician
  12 hours ago
  

  
  
  
  

 | 
show 9 more comments

$begingroup$

This question already has an answer here:

• 
  Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.
  
  13 answers
  
  

Let $G$ be a group. Show that if every non-identity element in $G$ has order $2$ then $G$ is abelian.

Proof:

Let $a,b $ be non-identity elements in $G$. Since $|a|=|b|=2$ , that means $ab=babaab$ $=$ $ba$.

Is the proof correct? How can I improve it?

abstract-algebra group-theory proof-verification abelian-groups

share|cite|improve this question

edited 2 days ago

TheSimpliFire

13.9k62765

asked May 4 at 21:03

topologicalmagiciantopologicalmagician

3299

$endgroup$

share|cite|improve this question

edited 2 days ago

TheSimpliFire

13.9k62765

asked May 4 at 21:03

topologicalmagiciantopologicalmagician

3299

share|cite|improve this question

edited 2 days ago

TheSimpliFire

13.9k62765

asked May 4 at 21:03

topologicalmagiciantopologicalmagician

3299

edited 2 days ago

TheSimpliFire

13.9k62765

edited 2 days ago

TheSimpliFire

13.9k62765

asked May 4 at 21:03

topologicalmagiciantopologicalmagician

3299

asked May 4 at 21:03

topologicalmagiciantopologicalmagician

3299

2 Answers
2

active

oldest

votes

4

$begingroup$

Alternative proof : $a^2=1$ so every element is its one inverse.

So, $(ab)^-1=ab$, but $(ab)^-1=b^-1a^-1=ba$ by using twice again the remark.

share|cite|improve this answer

answered May 4 at 21:19

elidiotelidiot

1,28014

$endgroup$

add a comment | 

4

$begingroup$

Your proof is correct.

You can improve it by making the derivations clearer (as you have done in the comments).

share|cite|improve this answer

answered May 4 at 21:14

ShaunShaun

11.2k113688

$endgroup$

add a comment | 

4

$begingroup$

Alternative proof : $a^2=1$ so every element is its one inverse.

So, $(ab)^-1=ab$, but $(ab)^-1=b^-1a^-1=ba$ by using twice again the remark.

share|cite|improve this answer

answered May 4 at 21:19

elidiotelidiot

1,28014

$endgroup$

add a comment | 

4

$begingroup$

Alternative proof : $a^2=1$ so every element is its one inverse.

So, $(ab)^-1=ab$, but $(ab)^-1=b^-1a^-1=ba$ by using twice again the remark.

share|cite|improve this answer

answered May 4 at 21:19

elidiotelidiot

1,28014

$endgroup$

add a comment | 

$begingroup$

Alternative proof : $a^2=1$ so every element is its one inverse.

So, $(ab)^-1=ab$, but $(ab)^-1=b^-1a^-1=ba$ by using twice again the remark.

share|cite|improve this answer

answered May 4 at 21:19

elidiotelidiot

1,28014

$endgroup$

share|cite|improve this answer

answered May 4 at 21:19

elidiotelidiot

1,28014

answered May 4 at 21:19

elidiotelidiot

1,28014

answered May 4 at 21:19

elidiotelidiot

1,28014

4

$begingroup$

Your proof is correct.

You can improve it by making the derivations clearer (as you have done in the comments).

share|cite|improve this answer

answered May 4 at 21:14

ShaunShaun

11.2k113688

$endgroup$

add a comment | 

4

$begingroup$

Your proof is correct.

You can improve it by making the derivations clearer (as you have done in the comments).

share|cite|improve this answer

answered May 4 at 21:14

ShaunShaun

11.2k113688

$endgroup$

add a comment | 

$begingroup$

Your proof is correct.

You can improve it by making the derivations clearer (as you have done in the comments).

share|cite|improve this answer

answered May 4 at 21:14

ShaunShaun

11.2k113688

$endgroup$

share|cite|improve this answer

answered May 4 at 21:14

ShaunShaun

11.2k113688

answered May 4 at 21:14

ShaunShaun

11.2k113688

answered May 4 at 21:14

ShaunShaun

11.2k113688