Let $G$ be a finite Abelian group of order $p^nm$, where $p$ is a prime that does not divide $m$. Then $G=H\times K$, where $H=\{x\in G|x^{p^n}=e\}$ and $K=\{x\in G|x^m=e\}$. Moreover,$|H|=p^n$.
I understand the part that $G=H\times K$. But I have problem understanding the part that $|H|=p^n$,Here is the prove from textbook
$p^nm=|HK|=\frac{|H||K|}{|H \cap K|}=|H||K|$,(because $G=H\times K$). It follows from following 2 theorems
1) In an Abelian group $G$, if $p$ divides $|G|$, then there exist an element in $G$ has order $p$.
2) if, $a^k=e$, then $|a|=k$.
that $p$ doesn't divide $|K|$. so $|H|=p^n$.
I don't understand how they get that $p$ doesn't divide $|K|$,suppose it is true, isn't this only tell us that $p^n$ divides $|H|$, why they two equal?

  • 2
    $\begingroup$ Your (2) is wrong as stated. The correct statement is that if $a^k=e$ then $|a|$ divides $k$. The reason that $p$ doesn't divide $|K|$ is that if it did, $K$ (by Cauchy's theorem) would have an element of order $p$, but since $x^m=e$ that would mean $p\mid m$, a contradiction. $\endgroup$ – anon Jul 6 '15 at 13:56
  • $\begingroup$ @anon what if I haven't learn Cauchy Theorem yet. $\endgroup$ – user236626 Jul 6 '15 at 14:09
  • 1
    $\begingroup$ Do you know subgroup of an Abelian group is abelian? If you know this , use your statement 1). $\endgroup$ – Chiranjeev_Kumar Jul 6 '15 at 14:15
  • $\begingroup$ Well, use [Cauchy's theorem for abelian groups]. That's what's stated in (1). $\endgroup$ – anon Jul 6 '15 at 17:45

If $p$ divides $|K|$, then by Cauchy Theorem $K$ will have an element of order $p$, Note that $H$ contains all the elements of order $p$ and integral power of $p$, This implies that $${H \cap K}\ne\phi$$

Which is not possible as $G=H\times K$, Hence $p$ does not divide $|K|$

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