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Group where every element is order 2

Let $(G,\star)$ be a group with identity element $e$ such that $a \star a = e$ for all $a \in G$. Prove that $G$ is abelian.

Ok, what i got is this: we want to prove that a*b=b*a, i.e. if a*a=e , a=a' where a' is the inverse and b*b=e, b=b' where b' is the inverse so a*b=(a*b)'=b'*a'=b*a....

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marked as duplicate by Arturo Magidin, Kannappan Sampath, Did, Sivaram Ambikasaran, Quixotic Mar 11 '12 at 7:51

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What does it mean to be abelian. Once you write out the definition can you see how to show a group is abelian? How does this relate to the hypotheses in the problem? –  john w. Mar 11 '12 at 4:05
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4 Answers 4

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HINT: You need to prove $ab = ba$, $\forall a,b \in G$. Note that since $ab \in G$ we also have that $(ab)^2 = e$ i.e. $(ab)(ab) = e$. Further, $a^{-1} = a$ and $b^{-1} = b$. Can you now finish this off?

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Ok, what i got is this: we want to prove that a*b=b*a, i.e. if a*a=e , a=a' where a' is the inverse and b*b=e, b=b' where b' is the inverse so a*b=(a*b)'=b'*a'=b*a.... –  anilorap Mar 11 '12 at 4:21
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If $a*a=e$ then $a=a^{-1}$. It follows that $(a*b)^{-1}=a*b$ but $a*b= b^{-1}*a^{-1}$ you get the desired equality for comparing both expressions.

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Given $a,b\in G$ we want to know $ab=ba$, i.e. $aba^{-1}b^{-1}=e$.

What we know is that $a*a=e$, i.e. $a=a^{-1}$, similarly $b=b^{-1}$, so what we need to verify is that $abab=e$.

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There are groups with $a\cdot a = e$ for all $a\in G$ with more than one element, so that reasoning will not work.

What you may want to think about: If $a\cdot a = e$ for all $a\in G$, then what does this tell you about the element $ab$? We know $(ab)^2 = e$ and so...

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