Prove that a group $G$ is abelian [duplicate]
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This question already has an answer here:
Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.
13 answers
Suppose we have a group $(G, *).$ Prove that the group is abelian if $b * a^2 = b$ where $(a, b)$ are part of the group.
abstract-algebra group-theory abelian-groups
edited Dec 1 '18 at 17:49
amWhy
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asked Dec 1 '18 at 17:38
David PriftiDavid Prifti
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Dec 1 '18 at 17:54
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.
add a comment |
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$begingroup$
This question already has an answer here:
Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.
13 answers
Suppose we have a group $(G, *).$ Prove that the group is abelian if $b * a^2 = b$ where $(a, b)$ are part of the group.
abstract-algebra group-theory abelian-groups
edited Dec 1 '18 at 17:49
amWhy
192k28225439
asked Dec 1 '18 at 17:38
David PriftiDavid Prifti
143
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marked as duplicate by Dietrich Burde abstract-algebra
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Dec 1 '18 at 17:54
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.
1
$begingroup$
What do you mean by $(a,b)$ are part of the group? Do you mean that $a,b in G$
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– Anurag A
Dec 1 '18 at 17:41
1
$begingroup$
this is not clear. Are you saying this should hold for all $a,bin G$? That only works if every element of $G$ has order $2$. Is that what you intended?
$endgroup$
– lulu
Dec 1 '18 at 17:42
$begingroup$
@AnuragA yes that is what I wanted to write
$endgroup$
– David Prifti
Dec 1 '18 at 17:47
$begingroup$
Once you cancel the $b$, you get $a^2=1$ for every $a$. Proving that this is abelian is a standard question, see for example math.stackexchange.com/questions/17054/…
$endgroup$
– verret
Dec 1 '18 at 17:48
$begingroup$
There are a couple ways but you should have already proven that inverses are unique and cancelation laws hold so $a^2*b = b$ means $a^2 = e$ and $a=a^{-1}$. and from ther $(ab)(ba) = ab^2a = a^2 =e$ so $ba = (ab)^{-1} = ab$.
$endgroup$
– fleablood
Dec 1 '18 at 17:58
add a comment |
0
0
0
$begingroup$
This question already has an answer here:
Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.
13 answers
Suppose we have a group $(G, *).$ Prove that the group is abelian if $b * a^2 = b$ where $(a, b)$ are part of the group.
abstract-algebra group-theory abelian-groups
edited Dec 1 '18 at 17:49
amWhy
192k28225439
asked Dec 1 '18 at 17:38
David PriftiDavid Prifti
143
$endgroup$
This question already has an answer here:
Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.
13 answers
Suppose we have a group $(G, *).$ Prove that the group is abelian if $b * a^2 = b$ where $(a, b)$ are part of the group.
This question already has an answer here:
Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.
13 answers
abstract-algebra group-theory abelian-groups
abstract-algebra group-theory abelian-groups
edited Dec 1 '18 at 17:49
amWhy
192k28225439
asked Dec 1 '18 at 17:38
David PriftiDavid Prifti
143
edited Dec 1 '18 at 17:49
amWhy
192k28225439
asked Dec 1 '18 at 17:38
David PriftiDavid Prifti
143
edited Dec 1 '18 at 17:49
amWhy
192k28225439
edited Dec 1 '18 at 17:49
amWhy
192k28225439
edited Dec 1 '18 at 17:49
amWhy
192k28225439
192k28225439
asked Dec 1 '18 at 17:38
David PriftiDavid Prifti
143
asked Dec 1 '18 at 17:38
David PriftiDavid Prifti
143
asked Dec 1 '18 at 17:38
David PriftiDavid Prifti
143
143
marked as duplicate by Dietrich Burde abstract-algebra
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Dec 1 '18 at 17:54
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Dec 1 '18 at 17:54
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.
1
$begingroup$
What do you mean by $(a,b)$ are part of the group? Do you mean that $a,b in G$
$endgroup$
– Anurag A
Dec 1 '18 at 17:41
1
$begingroup$
this is not clear. Are you saying this should hold for all $a,bin G$? That only works if every element of $G$ has order $2$. Is that what you intended?
$endgroup$
– lulu
Dec 1 '18 at 17:42
$begingroup$
@AnuragA yes that is what I wanted to write
$endgroup$
– David Prifti
Dec 1 '18 at 17:47
$begingroup$
Once you cancel the $b$, you get $a^2=1$ for every $a$. Proving that this is abelian is a standard question, see for example math.stackexchange.com/questions/17054/…
$endgroup$
– verret
Dec 1 '18 at 17:48
$begingroup$
There are a couple ways but you should have already proven that inverses are unique and cancelation laws hold so $a^2*b = b$ means $a^2 = e$ and $a=a^{-1}$. and from ther $(ab)(ba) = ab^2a = a^2 =e$ so $ba = (ab)^{-1} = ab$.
$endgroup$
– fleablood
Dec 1 '18 at 17:58
add a comment |
1
$begingroup$
What do you mean by $(a,b)$ are part of the group? Do you mean that $a,b in G$
$endgroup$
– Anurag A
Dec 1 '18 at 17:41
1
$begingroup$
this is not clear. Are you saying this should hold for all $a,bin G$? That only works if every element of $G$ has order $2$. Is that what you intended?
$endgroup$
– lulu
Dec 1 '18 at 17:42
$begingroup$
@AnuragA yes that is what I wanted to write
$endgroup$
– David Prifti
Dec 1 '18 at 17:47
$begingroup$
Once you cancel the $b$, you get $a^2=1$ for every $a$. Proving that this is abelian is a standard question, see for example math.stackexchange.com/questions/17054/…
$endgroup$
– verret
Dec 1 '18 at 17:48
$begingroup$
There are a couple ways but you should have already proven that inverses are unique and cancelation laws hold so $a^2*b = b$ means $a^2 = e$ and $a=a^{-1}$. and from ther $(ab)(ba) = ab^2a = a^2 =e$ so $ba = (ab)^{-1} = ab$.
$endgroup$
– fleablood
Dec 1 '18 at 17:58
1
1
$begingroup$
What do you mean by $(a,b)$ are part of the group? Do you mean that $a,b in G$
$endgroup$
– Anurag A
Dec 1 '18 at 17:41
$begingroup$
What do you mean by $(a,b)$ are part of the group? Do you mean that $a,b in G$
$endgroup$
– Anurag A
Dec 1 '18 at 17:41
1
1
$begingroup$
this is not clear. Are you saying this should hold for all $a,bin G$? That only works if every element of $G$ has order $2$. Is that what you intended?
$endgroup$
– lulu
Dec 1 '18 at 17:42
$begingroup$
this is not clear. Are you saying this should hold for all $a,bin G$? That only works if every element of $G$ has order $2$. Is that what you intended?
$endgroup$
– lulu
Dec 1 '18 at 17:42
$begingroup$
@AnuragA yes that is what I wanted to write
$endgroup$
– David Prifti
Dec 1 '18 at 17:47
$begingroup$
@AnuragA yes that is what I wanted to write
$endgroup$
– David Prifti
Dec 1 '18 at 17:47
$begingroup$
Once you cancel the $b$, you get $a^2=1$ for every $a$. Proving that this is abelian is a standard question, see for example math.stackexchange.com/questions/17054/…
$endgroup$
– verret
Dec 1 '18 at 17:48
$begingroup$
Once you cancel the $b$, you get $a^2=1$ for every $a$. Proving that this is abelian is a standard question, see for example math.stackexchange.com/questions/17054/…
$endgroup$
– verret
Dec 1 '18 at 17:48
$begingroup$
There are a couple ways but you should have already proven that inverses are unique and cancelation laws hold so $a^2*b = b$ means $a^2 = e$ and $a=a^{-1}$. and from ther $(ab)(ba) = ab^2a = a^2 =e$ so $ba = (ab)^{-1} = ab$.
$endgroup$
– fleablood
Dec 1 '18 at 17:58
$begingroup$
There are a couple ways but you should have already proven that inverses are unique and cancelation laws hold so $a^2*b = b$ means $a^2 = e$ and $a=a^{-1}$. and from ther $(ab)(ba) = ab^2a = a^2 =e$ so $ba = (ab)^{-1} = ab$.
$endgroup$
– fleablood
Dec 1 '18 at 17:58
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
Use the associative property to write
begin{align*}
ab &= (b^2a)(a^2b) \
&= b(ba^2)ab \
&= (ba)(ab) \
&= b(a^2b) \
&= ba.
end{align*}
Hence $G$ is Abelian.
answered Dec 1 '18 at 17:53
Sean RobersonSean Roberson
6,40031327
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Use the associative property to write
begin{align*}
ab &= (b^2a)(a^2b) \
&= b(ba^2)ab \
&= (ba)(ab) \
&= b(a^2b) \
&= ba.
end{align*}
Hence $G$ is Abelian.
answered Dec 1 '18 at 17:53
Sean RobersonSean Roberson
6,40031327
$endgroup$
add a comment |
1
$begingroup$
Use the associative property to write
begin{align*}
ab &= (b^2a)(a^2b) \
&= b(ba^2)ab \
&= (ba)(ab) \
&= b(a^2b) \
&= ba.
end{align*}
Hence $G$ is Abelian.
answered Dec 1 '18 at 17:53
Sean RobersonSean Roberson
6,40031327
$endgroup$
add a comment |
1
1
1
$begingroup$
Use the associative property to write
begin{align*}
ab &= (b^2a)(a^2b) \
&= b(ba^2)ab \
&= (ba)(ab) \
&= b(a^2b) \
&= ba.
end{align*}
Hence $G$ is Abelian.
answered Dec 1 '18 at 17:53
Sean RobersonSean Roberson
6,40031327
$endgroup$
Use the associative property to write
begin{align*}
ab &= (b^2a)(a^2b) \
&= b(ba^2)ab \
&= (ba)(ab) \
&= b(a^2b) \
&= ba.
end{align*}
Hence $G$ is Abelian.
answered Dec 1 '18 at 17:53
Sean RobersonSean Roberson
6,40031327
answered Dec 1 '18 at 17:53
Sean RobersonSean Roberson
6,40031327
answered Dec 1 '18 at 17:53
Sean RobersonSean Roberson
6,40031327
answered Dec 1 '18 at 17:53
Sean RobersonSean Roberson
6,40031327
6,40031327
add a comment |
add a comment |
1
$begingroup$
What do you mean by $(a,b)$ are part of the group? Do you mean that $a,b in G$
$endgroup$
– Anurag A
Dec 1 '18 at 17:41
1
$begingroup$
this is not clear. Are you saying this should hold for all $a,bin G$? That only works if every element of $G$ has order $2$. Is that what you intended?
$endgroup$
– lulu
Dec 1 '18 at 17:42
$begingroup$
@AnuragA yes that is what I wanted to write
$endgroup$
– David Prifti
Dec 1 '18 at 17:47
$begingroup$
Once you cancel the $b$, you get $a^2=1$ for every $a$. Proving that this is abelian is a standard question, see for example math.stackexchange.com/questions/17054/…
$endgroup$
– verret
Dec 1 '18 at 17:48
$begingroup$
There are a couple ways but you should have already proven that inverses are unique and cancelation laws hold so $a^2*b = b$ means $a^2 = e$ and $a=a^{-1}$. and from ther $(ab)(ba) = ab^2a = a^2 =e$ so $ba = (ab)^{-1} = ab$.
$endgroup$
– fleablood
Dec 1 '18 at 17:58