# Number of elements of order $n$ in an infinite group?

I tried to answer the following exercise:

Reformulate the corollary of Theorem 4.4. to include the case when the group has infinite order.

The corollary in question is this:

In a finite group the number of elements of order $n$ is divisble by $\varphi(n)$ where $\varphi$ is the totient function.

In an infinite group the number of elements of order $n$ is alse divisible by $\varphi(n)$.
• Let's regard the statement as being true when there are infinitely many elements of order $n$. If there are only finitely many elements in $G$ of order $n$, can you produce a finite collection of finite subgroups of $G$ containing all these elements? That way, you could make use of the result for when $G$ is finite. – D_S Nov 22 '15 at 1:19
Your answer doesn't make sense if the number of elements of order $n$ is infinite (or rather, your answer is vacuous in that case).
My guess is the following reformulation is the intended answer: If $G$ is a group, and $T_n$ is the set of elements of order $n$ in $G$, then the sets $\{g^k \mid \gcd(k, n)=1\}$ for $g \in T_n$ form a partition of $T_n$ into subsets of size $\varphi(n)$.