Number of elements of order $p$ is a multiple of $p-1$ (finite group).

This is a question in Pinter's A Book of Abstract Algebra.

Let $S=\{g\in G\mid \operatorname{ord}(g)=p\}$. Prove the order of $S$ is a multiple of $p-1$.

In his solution Pinter says $a \in S$ implies that $a$ generates a subgroup with $p-1$ elements. Shouldn't there be $p$ elements $\{1,a^1,\dots,a^{p-1}\}$? Or is it typical to only count the non-trivial elements in a subgroup?

Solution 1:

No, you are right, every element $a \in S$ generates a subgroup with $p$ elements. However, only $p-1$ of those will lie in the set $S$, which I guess is what Pinter means.

Solution 2:

You are correct, an element of order $n$ generates a subgroup with $n$ elements. Perhaps Pinter means that the subgroup generated by an element of order $p$ contains $p-1$ elements of order $p$ (namely, the nontrivial ones).