Abelian Group of Prime-power Order is Product of Cyclic Groups

Lemma
Let $$G$$ be an abelian group of prime-power order.

Let $$a$$ be an element of maximal order in $$G$$.

Then $$G$$ can be written in the form $$\left \langle {a} \right \rangle \times K$$.

Proof
Denote $$\left|{G}\right| = p^n$$ and induct on $$n$$.

For $$n=1$$, then $$G = \langle a \rangle \times \langle e \rangle$$.

Now we assume the lemma is true for all Abelian groups of order $$p^k$$, where $$k < n$$.

For an element $$a \in G$$ with maximal order $$p^m, x^{p^m} = e \forall x \in G$$.

We can assume $$G \ne \langle a \rangle$$, for then there is nothing to prove.

Choose $$b \in G$$ such that $$b \notin \langle a \rangle$$.

The claim is that $$\langle a \rangle \cap \langle b \rangle = \left\{ e \right\}$$.

Clearly, this would be established if $$\left|{b}\right| = p$$.

Since $$\left| {b^p}\right| = \frac {\left|{b}\right|} p$$, we know that $$b^p \in \langle a \rangle$$ by the manner in which b was chosen.

Say $$b^p = a^i$$.

Note that $$e = b^{p^m} = \left({b^p}\right)^{p^{m-1}} = \left({a^i}\right)^{p^{m-1}}$$, so $$\left|a^i\right|\le p^{m-1}$$.

Hence $$a^i$$ is not a generator of $$\langle a \rangle$$ and therefore $$gcd \left\{{p^m, i}\right\} \ne 1$$.

This proves that $$p$$ divides $$i$$, so that we can write $$i=pj$$.

Then $$b^p = a^i = a^{pj}$$.

Consider that element $$c = a^{-j} b$$.

This element $$c$$ is not in $$\langle a \rangle$$, for if it were, $$b$$ would be, too.

Also, $$c^p = a^{-jp} b^p = a^{-i}b^p = b^{-p}b^p = e$$.

Hence $$c$$ is an element of order $$p$$ such that $$c \notin \langle a \rangle$$.

Since $$b$$ was chosen to have smallest order such that $$b \notin \langle a \rangle$$, we conclude $$b$$ also has order $$p$$, and the claim is verified.

Now consider the factor group $$\bar{G} = G / \langle b \rangle$$.

Let $$\bar{x}$$ denote the coset $$x \langle b \rangle$$ in G.

If $$\left|\bar{a}\right| < \left|a\right| = p^m$$, then $$\bar{a}^{p^{m-1}} = \bar{e}$$.

This means $$\left({a \langle b \rangle}\right)^{p^{m-1}} = a^{p^{m-1}} \langle b \rangle = \langle b \rangle$$, so that $$a^{p^{m-1}} \in \langle a \rangle \cap \langle b \rangle = \left\{{e}\right\}$$, contradicting the fact that $$\left|a\right| = p^m$$.

Thus, $$\left|\bar{a}\right| = \left|a\right| = p^m$$, and therefore $$\bar{a}$$ is an element of maximal order in $$\bar{G}$$.

By induction, we know that $$\bar{G}$$ can be written in the form $$\langle \bar{a} \rangle \times \bar{K}$$ for some subgroup $$\bar{K}$$ of $$\bar{G}$$.

Let $$K$$ be the pullback of $$\bar{K}$$ under the natural homomorphism from $$G$$ to $$\bar{G}$$, specifically, $$K = \left\{{x \in G : \bar{x} \in \bar{K}}\right\}$$.

The claim is that $$\langle a \rangle \cap K = \left\{{e}\right\}$$.

For if $$x \in \langle a \rangle \cap K$$, then $$\bar{x} \in \langle \bar{a} \rangle \cap \bar{K} = \left\{{\bar{e}}\right\} = \langle b \rangle$$ and $$x \in \langle a \rangle \cap \langle b \rangle = \left\{ e \right\}$$.

It now follows from an order argument that $$G = \langle a \rangle K$$, and therefore $$G = \langle a \rangle \times K$$.