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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?

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    Hint: What is the order of 1?2011-12-18
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    The unit element has order $1$.2011-12-18
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    I suppose that $G$ is a group and $p$ is a prime number. You should include such information into your question.2011-12-18
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    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.2011-12-18
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    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.2011-12-18

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