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Problem

Let $G$ be a finite abelian group and $p$ a positive prime that divides $|G|$. Show that the number of elements of order $p$ in $G$ is coprime with $p$.

Let $|G|=p^nm$ with $n \geq 1$ and $gcd(p,m)=1$. If we consider $G$ as a $\mathbb Z$-module, then by the structure theorem we have $$G \cong \mathbb Z_{p^{n_1}} \oplus ... \oplus \mathbb Z_{p^{n_k}} \oplus M$$ with $\mathbb Z_{p^j} \not \subset M$ for all $j \in \mathbb N$ and with $n_1+...+n_k=n$.

I don't know what to do, I got stuck here, I would appreciate some help with this problem. Thanks in advance.

user156441
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  • It is always true,($G$ is abelian or not). Actually if $k$ is the number of the elements of order $p$ then $k\equiv -1 \mod \ p$ – mesel Dec 08 '14 at 21:28
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    Hint: Since $G$ is Abelian, the elements of order dividing $p$ form a subgroup, which has order $p^{m}$ for some positive integer $m.$ – Geoff Robinson Dec 08 '14 at 21:30
  • @GeoffRobinson: So being abelian is for easy conclusion ? – mesel Dec 08 '14 at 21:34
  • @mesel : It is easier to prove in the Abelian case( which is the case of the question). As you say, it is also true for non-Abelian $G.$ – Geoff Robinson Dec 08 '14 at 21:36
  • @GeoffRobinson Thanks for the hint. If $m=1$, then $n_p=p-1$ which is coprime with p. If $m=2$, then by Euler's totient function, there are $\phi(p^2)=p^2-p$ elements of order $p^2$, so there are $p^2-\phi(p^2)-1=p-1$ elements again. But what about $m > 2$? – user156441 Dec 08 '14 at 21:38
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    The method you are using does not follow the hint I gave. You only need to worry about elements of order $1$ or $p$ if you use the hint. – Geoff Robinson Dec 08 '14 at 21:42
  • Oh, you're absolutely right, there are $p^m-1$ elements of order $p$, which is coprime with $p$. Thank you! – user156441 Dec 08 '14 at 21:44

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As pointed out in the comments, the claim holds in general. The $N_p$ elements of order $p$ are grouped into $\ell_p$ trivially intersecting subgroups of order $p$: $$N_p=\ell_p(p-1)\equiv -\ell_p\pmod p$$ But, as a corollary of Wielandt's proof of Sylow's theorem, $\ell_p\equiv 1\pmod p$. Therefore $$N_p\equiv -1\pmod p$$ and hence $\gcd(N_p,p)=1$.