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
First note that from Lemma 1, both $H$ and $K$ are subgroups of $G$.

Because $G$ is abelian, $H$ and $K$ are normal, so from the Internal Direct Product Theorem, 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 $a^{\left|{G}\right|} = e$ from Element to the Power of Group Order.

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 $x^{p^n} = e = x^m$, so from Element to the Power of Multiple of Order $\left|{x}\right|$ divides both $p^n$ and $m$.

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

Suppose that $p \backslash \left|{K}\right|$.

From Cauchy's Theorem for Abelian Groups, this implies that there is some element (call it $k$) of $K$ with order $p$.

But we also have $k^m = e$ from the definition of $K$, so by Element to the Power of Multiple of Order we must have $p \backslash m$, a contradiction, so our supposition must be false and $p \nmid \left|{K}\right|$.

Now suppose a prime other than $p$ divides $H$. Call this prime $q$.

Again from Cauchy's Theorem for Abelian Groups. This implies that there is some element $h$ of $H$ with order $q$.

But since $h^{p^n} = e$ from the definition of $H$, Element to the Power of Multiple of Order gives us $q \backslash p^n$, a contradiction, so our supposition must be false and $q \nmid \left|{H}\right|$.

It then follows that $\left|{H}\right| = p^n$ as desired.