Abelian Group Factored by Prime

Theorem
Let $G$ be a finite abelian group and let $p$ be a prime.

Factor the order of $G$ as $\left|{G}\right| = m p^n$ such that $p$ does not divide $m$.

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

Furthermore, we have $\left|{H}\right| = p^n$.

Corollary
Any finite abelian group $G$ can be factored as follows:

Let $\left|{G}\right| = \displaystyle \prod_{i=1}^k p_i^{n_i}$ be the prime factorisation of the order of $G$.

Then we have $G = \displaystyle \prod_{i=1}^k H_i$, where $H_i = \left\{{x \in G : x^{p_i^{n_i}} = e}\right\}$.

This factorisation is unique up to ordering of the factors.

Proof
First note that from Subgroup of Elements whose Order Divides Integer, both $H$ and $K$ are subgroups of $G$.

Also, because $G$ is abelian, $H$ and $K$ are normal by All Subgroups of Abelian Group are Normal.

In order to prove $G = H \times K$, by the Internal Direct Product Theorem it suffices to 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 $1 = s m + t p^n$ by Bézout's Lemma.

Let $s$ and $t$ be two such integers.

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

Now we have $x^{\left|{G}\right|} = e$ from Element to the Power of Group Order.

Therefore, $(x^{s m})^{p^n} = (x^{p^n m})^s = e^s = e$ and $(x^{t p^n})^m = (x^{p^n m})^t = e^t = e$.

By definition, $x^{s m} \in H$ and $x^{t p^n}\in K$.

We conclude that $G = H K$.

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 $m$, it follows that $\left|{x}\right| = 1$.

Therefore, by Identity Only Group Element Order 1, $x = e$, thus $H \cap K = \left\{e\right\}$.

It follows that $G = H \times K$.

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

From Cauchy's Group Theorem, 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.

We conclude that $p$ does not divide $\left|{K}\right|$, and it follows that $p^n \backslash \left|{H}\right|$.

Now suppose a prime $q$ (with $q \ne p$) divides $H$.

Again from Cauchy's Group Theorem, 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.

It follows that $q$ does not divide $\left|{H}\right|$.

We conclude $\left|{H}\right| = p^n$, as desired.

Proof of Corollary
Suppose $\left|{G}\right| = \displaystyle \prod_{i=1}^k p_i^{n_i}$ is the prime factorisation of $\left|{G}\right|$.

We proceed by induction on $k$.

Basis for the induction
For $n=1$, the statement is trivial.

Induction Hypothesis
Now we assume the theorem is true for abelian groups whose order has $k-1$ distinct prime factors.

Induction Step
Apply the theorem to $G$ and $p_1$. By definition, $H = H_1$.

Also. the resulting $K$ has $\left|{K}\right| = \displaystyle \prod_{i=2}^k p_i^{n_i}$.

Therefore, it satisfies the induction hypothesis.

It follows that $G = H_1 \times \displaystyle \displaystyle \prod_{i=2}^k H_i$.

Furthermore, all the $H_i$ are normal Sylow $p$-subgroups.

From Normal Sylow P-Subgroup is Unique, the factorisation is unique up to ordering of the factors.