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?

  • 3
    $\begingroup$ You even say "Then since $G$ is abelian...". $\endgroup$ – Dan Rust Dec 16 '14 at 11:33
  • $\begingroup$ Did not realize the property $(ab)^n=a^nb^n$ only if group is abelian $\endgroup$ – Sam Houston Dec 16 '14 at 11:39

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 $\Bbb Z_2 \star \Bbb Z_2 = \langle a, b \mid a^2 = b^2 = 1 \rangle$

  • $\begingroup$ Thanks for your answer, but why does that not hold? Can you direct me to where I can read why this is the case. $\endgroup$ – Sam Houston Dec 16 '14 at 11:33
  • $\begingroup$ no worries found it $\endgroup$ – Sam Houston Dec 16 '14 at 11:34
  • $\begingroup$ There, @Dansmith, I gave you an example. $\endgroup$ – Balarka Sen Dec 16 '14 at 11:36
  • $\begingroup$ Thanks, just what I was looking for $\endgroup$ – Sam Houston Dec 16 '14 at 11:37

The last step is problematic: Consider the permutations $a=(1,2)$ and $b=(2,3)$. Then $a^2b^2$ is trivial, but $ab=(1,3,2)$ has order 3 and so $(ab)^2\neq a^2 b^2$.

And of course this can even extend to finite vs. infinite order, as you can see in, say, a group with presentation $\langle x,y\mid x^2=y^2=1 \rangle$. There, $x$ and $y$ have order 2, but $xy$ has infinite order.


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