Abelian Group Factored by Prime

Lemma
Let $$G$$ be a finite abelian group of order $m p^n$ where $$p$$ is a prime that does not divide $$m$$.

Then $$G = H \times K$$ where $$H = \left\{{x \in G : x^{p^n} = e}\right\}$$ and $$K = \left\{{x \in G : x^m = e}\right\}$$.

$$\left|{H}\right| = p^n$$.

Proof
Because $$G$$ is abelian, to prove $$G = H \times K$$ we need only show that $$G = H K$$ and $$H \cap K = \left\{e\right\}$$.

Since we have $$\gcd \left\{{m, p^n}\right\} = 1$$, there are integers $$s$$ and $$t$$ such that $$1 = s m + t p^n$$.

So it follows that $$\forall x \in G: x = x^{s m + t p^n}$$

Now we have by Lagrange's Theorem that $$\forall a \in G: \left|{a}\right| \backslash \left|{G}\right|$$, which implies $$a^{\left|{G}\right|} = e$$.

So $$x^{sm} \in H$$ and $$x^{tp^n}\in K$$.

Thus $$G = H K$$.

Now suppose that some $$x \in H \cap K$$.

Then $$x^{p^n} = e = x^m$$, and hence $$\left|{x}\right|$$ divides both $$p^n$$ and $$m$$.

Since $$p$$ does not divide $$m$$, $$\left|{x}\right| = 1 \implies x = e$$.