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Let $G$ be a finite abelian group of order $p^nm$ , where $p$ is prime that does not divide $m$. Then show that $G = H\times K$ , where $H = \{x\in G | x^{p^n} = e \}$ and $K= \{x\in G | x^m= e\}$.

Kenta S
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Define $t:G\rightarrow H\times K$ by $t(x)=(x^m,x^{p^n})$. It is easy to verify that $t$ is a homomorphism. To complete the proof, you must show this homomorphism is bijective, which involves a bit of modulo arithmetic.

tjeremie
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