TITLE: Prove that a group is abelian.
QUESTION [8 upvotes]: Possible Duplicate:
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 ab=ba, i.e. if aa=e , a=a' where a' is the inverse and bb=e, b=b' where b' is the inverse so ab=(ab)'=b'a'=ba....

REPLY [10 votes]: 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?

REPLY [3 votes]: 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 $abab=e$.

REPLY [2 votes]: 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.