TITLE: No. of finite group (nonidentity)elements $x$ satisfying $x^5=e$ is a multiple of $4$
QUESTION [7 upvotes]: In a finite group $G$ with $e:=\text{id}_G$, show that the number of nonidentity elements that satisfy the equation $x^5=e$ is a multiple of 4.  

This is number $50$, Ch.$2$ from Gallian's text. I have seen two repeats of this question on MSE (Show that number of solutions satisfying $x^5=e$ is a multiple of 4?, In a finite group, show that the number of nonidentity elements that satisfy the equation $x^5=e$ is a multiple of 4.) but I still have questions about the question and my proof. 
My first question: 
Do we have to assume that $x^5=e$ for some $x\in G$ in the first place? My thinking is that this is a yes since in general a finite group may not have such an $x$. 
Proof of claim:
Suppose some $x\in G$,$\space$$x\neq e$ satisfies the condition $x^5=e$. Then note that $x^2\in G$ and $$(x^2)^5=(x^5)^2=e^2=e$$ so $x^2$ satisfies the condition. Similarly $x^3\in G$ and $x^4\in G$. Observe that $$(x^3)^5=(x^5)^3=e^3=e$$ $$(x^4)^5=(x^5)^4=e^4=e$$ so that $x^3$ and $x^4$ also satisfy the condition. Once we verify that $x,x^2,x^3,x^4$ are distinct and that $x^i\neq e$ for $1\leq i \leq 4$ we will have proved the claim since for every $x$ that is a solution, so is $x^2$,$x^3$, and $x^4$. Thus solutions come in multiples of $4$. 
Note that we don't consider elements like $x^6$ or $x^7$ as solutions because $x^6=x$ and $x^7=x^2$ i.e. for $n>5$, $x^n=x^i$ where $i\in\{1,2,3,4,5\}$. So, considering powers of elements modulo $5$ is enough.
To show each $x^i$ is distinct, we assume the contrary. That is, assume $$x^i=x^j$$ for some distinct $i,j\in\{1,2,3,4\}$. Thus if we take $i$ to always be the greater of the two, $$x^i=x^j\iff x^{i-j}=e$$ so $$i-j=1,2\text{ or } 3$$
Note it is impossible $i-j=1$ by the assumption that $x\neq e$. If $i-j=2$ or $i-j=3$, then $$x^2=e\text{ and } x^3=e\implies x^3=x^2\cdot x=e\cdot x=x = e$$
but the latter shows $x=e$ if $x^3=e$ and $x^2=e$; this is a contradiciton. Thus it must be false that $x^i=x^j$ for some distinct $i,j\in\{1,2,3,4,5\}$. 
In proving the above, we saw that the $x^2=e=x^3$ leads to a contradiction, so to show the last claim, namely that $x^i\neq e$, we show that $x^4\neq e$. Again assume that indeed $x^4=e$. Then $$x^4=e=x^5\iff e=x\therefore\text{ contradiction }$$ $\blacksquare$
My second question: 
Is the above proof correct? By removing the condition that the group be finite, how could this change the conclusion about the number of solutions? I never really utilized that $G$ was finite above (maybe tacitly? I don't know) so I'm pretty stumped on this one.

REPLY [3 votes]: In general, if $p$ is an odd prime and $G$ a finite group, then $\#\{ g \in G: g^p=1\} \equiv 1$ mod $(p-1)$. Observe that the set includes the identity element. Proof (sketch): on the set $S=\{ g \in G: g^p=1\}$ define an equivalence relation: $g \sim h$ if and only if $\langle g \rangle =\langle h \rangle$. Then $S$ partitions in $\{1\}$ and equivalence classes of order $p-1$ (namely $\langle g \rangle -\{1\}$ for each non-identity $g \in S$).