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?
Best Answer
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.