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 $\forall x \in G: x = x^{s m + t p^n} = x^{sm}x^{tp^n}$.

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 $(x^{sm})^{p^n} = (x^{p^nm})^s = e^s = e$ and $(x^{tp^n})^m = (x^{p^nm})^t = e^t = e$.

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

Therefore $G = H K$.

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 $m$, $\left|{x}\right| = 1 \implies x = e$.