Suppose we're given an infinite group $G$ with no additional structure. How many elements of that group have order $n$? There is a related question here, but I was surprised to see this not mentioned.
For finite groups, we have the counting theorem:
In a finite group, the number of elements of order $d$ is a multiple of $\phi(d)$.
The proof doesn't really rely on the fact that $G$ is finite, but does only insofar as to say the multiple of $\phi(d)$ "exists" as a finite number.
Isn't it legitimate to say that in infinite groups, there are either a finite number of elements of order $d$, namely $k\cdot\phi(d)$ for some nonnegative integer $k$, or an infinite number of such elements (infinitely many multiples of $\phi(d)$) ?
Let $G$ be any group and fix some integer $d > 0$. Suppose that only finitely many elements have order $d$, say $g_1, \ldots, g_n$. Let $m < n$ be the largest integer such that, after potentially reordering the $g_i$, then no power of an element of $g_1, \ldots, g_m$ is a power of another. That is, for $i, j < m$, $g_i^k = g_j^s\neq e$ implies that $i = j$.
Now consider the (necessarily finite) cyclic groups $\langle g_1\rangle, \ldots \langle g_m\rangle$. Each $g_i$ has $\phi(d)$ generators, $g_i^{k_1}, \ldots, g_i^{k_{\phi(d)}}$, where the $k_j$ are the integers $< d$ that are relatively prime to $d$. By assumption, this gives $\phi(d)m$ distinct elements of order $d$. Now suppose that some $g_r$ (for $r > m$) doesn't show up as one of these generators. If $g_i^t = g_r^s \neq e$ for some $i \leq m$, then $g_i^{t+d-s+1} = g_r$. Since $g_r$ has order $d$, then $t + d - s + 1 \equiv k_j\pmod{d}$ for some $k_j \in \{k_1, \ldots, k_{\phi(d)}\}$. Therefore, $g_i^{k_j} = g_r$, which is a contradiction.
In other words, $g_1, \ldots, g_m, g_r$ then satisfies the condition that $g_i^r = g_j^s \neq e$ implies $i = j$, which contradicts maximality of $m$. Therefore, every element of order $d$ shows up as a (relatively prime to $d$) power of one of the $g_1, \ldots, g_m$. This shows that there are $\phi(d)m$ elements of order $d$.