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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.

amWhy
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David Prifti
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    What do you mean by $(a,b)$ are part of the group? Do you mean that $a,b \in G$ – Anurag A Dec 01 '18 at 17:41
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    this is not clear. Are you saying this should hold for all $a,b\in G$? That only works if every element of $G$ has order $2$. Is that what you intended? – lulu Dec 01 '18 at 17:42
  • @AnuragA yes that is what I wanted to write – David Prifti Dec 01 '18 at 17:47
  • 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 https://math.stackexchange.com/questions/17054/group-where-every-element-is-order-2 – verret Dec 01 '18 at 17:48
  • 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 $ba = (ab)^{-1} = ab$. – fleablood Dec 01 '18 at 17:58

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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.

Sean Roberson
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