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

Let $S=\{g\in G\mid \operatorname{order}(g)=p\}$; 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? $\{1,a^1,\cdots,a^{p-1}\}$ Or is it typical to only count the non-trivial elements in a subgroup?

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Hint: What is the order of 1? – Dylan Wilson Dec 18 '11 at 20:49
The unit element has order $1$. – user18119 Dec 18 '11 at 20:49
I suppose that $G$ is a group and $p$ is a prime number. You should include such information into your question. – Martin Sleziak Dec 18 '11 at 20:49
You're right in that $a$ will generate a subgroup of order $p$. I think the point is that the subgroup will have $p - 1$ elements of order $p$, and then the identity. So the elements of order $p$ come in $(p - 1)$-sized bunches. – Dylan Moreland Dec 18 '11 at 20:49
One way of looking at this is that $(\mathbf Z/p\mathbf Z)^*$ acts on $S$ by $n.x = x^n$, and the orbits of this action (which partition $S$) all have size $p - 1$, since the stabilizers are all trivial. – Dylan Moreland Dec 18 '11 at 21:07

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.

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Yes, I suppose that the sentence is to be read "every element in $S$ generates a subgroup that has $p-1$ elements in $S$". – Andrea Mori Dec 18 '11 at 23:15

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).

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