Number of elements in a Group such that $x^7=e$

Question

Given that $G$ is a finite Group. Prove that number of elements in G such that $x^7=e$ where $x \in G$ is always Odd.

My attempt:

First possibility is a Trivial group since $e^7=e$. Trivial Group contains One element which is Odd number.

Now let $x\ne e$,then since $x^7=e$ and $7$ is prime we have $ord(x)=7$ and this implies $ord(x^i)=7$ $\:$ $\forall 1\le i\le 6$

So minimum possible non trivial group is $$G=\left\{e, x,x^2, x^3,x^4,x^5,x^6\right\}$$ Which is a Cyclic Group containing $7$ elements which is Odd.

Next possibility is another Cyclic group with $13$ elements which is Odd. So in general the Cardinality of the group is of the form $6n+1$.

But are all these cyclic groups only? Does it mean if $ord(x)$ is Finite, then the group is Cyclic?

Answer 1

In general, if $p$ is an odd prime and $G$ a finite group, then $$\#\{ g \in G\mid g^p=e\} \equiv 1 \bmod (p-1).$$

Proof: See No. of finite group (nonidentity)elements $x$ satisfying $x^5=e$ is a multiple of $4$

Answer 2

There's another way :

Think about it. If a group $G$ is cyclic and order of an element $a$ is $7$ then $<a>$ is the only subgroup containing all the elements of order 7, which are $\phi {7} = 6 $ in number and also for Identity $e$ $e^7= e$ too. So there's only odd number of such elements $x$ in $G$.

If a group is not a cyclic group, then if a be an element of order 7 then $<a>$ contains 6 elements of order $7$. If this exhausts all order $7$ elements then we have $6+1=7$ i.e odd number of such elements $x$.

If there's another element $b$ of order 7 which doesn't belong to $<a>$ then $<b> \cap <a> = \phi$. Thus the $6$ elements of $<b>$ i.e. $\{b, b^2,....,b^6 \}$ are distinct elements of order 7 and now the total number of order 7 is $2\phi(7) = 12$. If this process exhausts all the elements of order $7$ then we are done otherwise going with the same argument we have that the number of order 7 elements in this group is a multiple of $\phi(7)$ i.e. $n\phi(7) = 6n$ for some integer $n$.

Therefore the number of such $x$ is $6n+1$, which is odd.

So you can see that if $|x|=7$ then it's not necessary that the group must be cyclic.