If in a group $G$ , $(ab)^2=(ba)^2$ for all $a,b in G$ , then show that $G$ is abelian .
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Initially , I have placed $a^-1b$ in place of $b$ , and obtained $a^2b= ba^2$ for all $ain G$. Consequently, i have obtained $ab^2=b^2a$ . Then what to do ?
group-theory
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Initially , I have placed $a^-1b$ in place of $b$ , and obtained $a^2b= ba^2$ for all $ain G$. Consequently, i have obtained $ab^2=b^2a$ . Then what to do ?
group-theory
Perhaps that you meant $(ab)^2=a^2b^2$.
â José Carlos Santos
Jul 22 at 17:27
3
It is true under additional hypotheses, compare $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$ or Prove that a ring is commutative if $(ab)^2=(ba)^2$
â Martin R
Jul 22 at 17:29
1
The quaternion group does obey the law $ab^2=b^2a$.
â Lord Shark the Unknown
Jul 22 at 17:31
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Initially , I have placed $a^-1b$ in place of $b$ , and obtained $a^2b= ba^2$ for all $ain G$. Consequently, i have obtained $ab^2=b^2a$ . Then what to do ?
group-theory
Initially , I have placed $a^-1b$ in place of $b$ , and obtained $a^2b= ba^2$ for all $ain G$. Consequently, i have obtained $ab^2=b^2a$ . Then what to do ?
group-theory
edited Jul 22 at 18:13
Shaun
7,33892972
7,33892972
asked Jul 22 at 17:20
RABI KUMAR CHAKRABORTY
244
244
Perhaps that you meant $(ab)^2=a^2b^2$.
â José Carlos Santos
Jul 22 at 17:27
3
It is true under additional hypotheses, compare $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$ or Prove that a ring is commutative if $(ab)^2=(ba)^2$
â Martin R
Jul 22 at 17:29
1
The quaternion group does obey the law $ab^2=b^2a$.
â Lord Shark the Unknown
Jul 22 at 17:31
add a comment |Â
Perhaps that you meant $(ab)^2=a^2b^2$.
â José Carlos Santos
Jul 22 at 17:27
3
It is true under additional hypotheses, compare $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$ or Prove that a ring is commutative if $(ab)^2=(ba)^2$
â Martin R
Jul 22 at 17:29
1
The quaternion group does obey the law $ab^2=b^2a$.
â Lord Shark the Unknown
Jul 22 at 17:31
Perhaps that you meant $(ab)^2=a^2b^2$.
â José Carlos Santos
Jul 22 at 17:27
Perhaps that you meant $(ab)^2=a^2b^2$.
â José Carlos Santos
Jul 22 at 17:27
3
3
It is true under additional hypotheses, compare $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$ or Prove that a ring is commutative if $(ab)^2=(ba)^2$
â Martin R
Jul 22 at 17:29
It is true under additional hypotheses, compare $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$ or Prove that a ring is commutative if $(ab)^2=(ba)^2$
â Martin R
Jul 22 at 17:29
1
1
The quaternion group does obey the law $ab^2=b^2a$.
â Lord Shark the Unknown
Jul 22 at 17:31
The quaternion group does obey the law $ab^2=b^2a$.
â Lord Shark the Unknown
Jul 22 at 17:31
add a comment |Â
1 Answer
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You don't. A counterexample is the quaternion group of order $8$.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
You don't. A counterexample is the quaternion group of order $8$.
add a comment |Â
up vote
11
down vote
You don't. A counterexample is the quaternion group of order $8$.
add a comment |Â
up vote
11
down vote
up vote
11
down vote
You don't. A counterexample is the quaternion group of order $8$.
You don't. A counterexample is the quaternion group of order $8$.
answered Jul 22 at 17:26
Lord Shark the Unknown
85.2k950111
85.2k950111
add a comment |Â
add a comment |Â
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Perhaps that you meant $(ab)^2=a^2b^2$.
â José Carlos Santos
Jul 22 at 17:27
3
It is true under additional hypotheses, compare $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$ or Prove that a ring is commutative if $(ab)^2=(ba)^2$
â Martin R
Jul 22 at 17:29
1
The quaternion group does obey the law $ab^2=b^2a$.
â Lord Shark the Unknown
Jul 22 at 17:31