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 $K= \{x\in G | x^m= e\}$.
Asked
Active
Viewed 36 times
1 Answers
0
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
- 712