Let G be a group. Show that if $G$ is abelian then the set of elements in $G$ of finite order form a subgroup.
I have a proof for this question but I dont understand how the group has to be abelian for the implication.
Let $H$ be the set of elements of $G$ of finite order.
The identity element $e$ has order $1$, so $e\in H$.
If $a\in H$ then $\langle a^{-1}\rangle=\langle a\rangle$, so $a^{-1}$ has the same order as $a$, and hence $a^{-1}\in H$.
Let $a,b\in H$, where $a$ has order $m$ and $b$ has order $n$.
Then since $G$ is abelian, $(ab)^{mn}=a^{mn}b^{mn}=e$
I understand each step, but don't see how having $G$ being an abelian group is necessary for any of these steps. To me they appear to hold regardless of $G$ being abelian or not. Which step of the proof would the abelian property be required? Why?
Best Answer
If $G$ is not abelian, $(ab)^n = a^nb^n$ might not hold. So even if $a$ and $b$ have finite order, $ab$ might or might not have finite order. Thus the collection of elements having having finite order in $G$ might not be closed under multiplication.
An example of a nonabelian group with multiplication not preserving the finite order property is the free product $\Bbb Z_2 \star \Bbb Z_2 = \langle a, b \mid a^2 = b^2 = 1 \rangle$