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

Theorem
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 $\left\langle{a}\right\rangle \times K$ for some $K \le G$.

Proof
Suppose $\left|{G}\right| = p^n$ with $p$ a prime.

We proceed by induction on $n$.

Basis for the induction
For $n=1$, we have $G = \left\langle{a}\right\rangle \times \left\langle{e}\right\rangle$, by Group of Prime Order is Cyclic.

Induction Hypothesis
Now we assume the theorem is true for all abelian groups of order $p^k$, where $k < n$.

Induction Step
Let $a$ be an element with maximal order $p^m$ in $G$.

For any $x \in G$, we have $\left|{x}\right| \backslash \left|{G}\right|$ by Order of Element Divides Order of Finite Group.

Thus we have $\left|{x}\right| = p^i$, for some $i \geq 0$. By assumption, $i \le m$.

We conclude that $\forall x \in G : x^{p^m} = e$.

Now assume $G \ne \left\langle{a}\right\rangle$, for the remaining case is trivial.

Choose $b \in G$ such that $\left|{c}\right| < \left|{b}\right|$ implies $c \in \left\langle{a}\right\rangle$, but $b \not \in \left\langle{a}\right\rangle$.

Lemma
We have $\left\langle{a}\right\rangle \cap \left\langle{b}\right\rangle = \left\langle{e}\right\rangle$.

 Since $\left|{b^p}\right| = \dfrac {\left|{b}\right|} p < \left|{b}\right|$, from the definition of $b$ we conclude $b^p \in \left\langle{a}\right\rangle$.

Thus we have $\exists i \in \Z : b^p = a^i$.

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 $\left\langle{a}\right\rangle$ and therefore $\gcd \left\{{p^m, i}\right\} \ne 1$.

It follows that $p$ divides $i$, so that we can write $i = pj$ for a $j \in \Z$.

This means that we have $b^p = a^i = a^{pj}$.

Now consider the element $c = a^{-j} b$.

This element $c$ is not in $\left\langle{a}\right\rangle$, for if it were, $b = a^j c$ would be as well.

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

Hence, $\left|{c}\right| = p$, and $c \notin \langle a \rangle$.

From the definition of $b$, we conclude that also $\left|{b}\right| = p$.

It follows from Group of Prime Order is Cyclic that $\left\langle{b}\right\rangle$ is generated by all its non-identity elements.

Therefore, by Intersection of Subgroups, $\left\langle{a}\right\rangle \cap \left\langle{b}\right\rangle$ is either $\left\langle{b}\right\rangle$ or $\left\langle{e}\right\rangle$.

However, the first case contradicts $b \not \in \left\langle{a}\right\rangle$.

We conclude that $\left\langle{a}\right\rangle \cap \left\langle{b}\right\rangle = \left\langle{e}\right\rangle$.

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

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

Assume that $\left|\bar a\right| < \left|a\right| = p^m$.

We observe that therefore $\bar a^{p^{m-1}} = \bar e$.

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

This contradicts $\left|{a}\right| = p^m$.

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

As $\left|{\bar G}\right| = p^{n-1}$ by a corollary of Quotient Group is Group, the induction hypothesis applies.

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

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

As this natural homomorphism is $p$-to-1 (i.e., every element in the image has $p$ preimages), it follows that $\left|{K}\right| = p \left|{\bar K}\right|$.

Also, Pullback is Subgroup ensures that $K \leq G$.

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

It follows that $x \in \left\langle{a}\right\rangle \cap \left\langle{b}\right\rangle = \left\langle{e}\right\rangle$ and therefore, $\left\langle{a}\right\rangle \cap K = \left\langle{e}\right\rangle$.

This means that $\left|{ \left\langle{a}\right\rangle K }\right| = \left|{ \left\langle{a}\right\rangle }\right| \left|{ K }\right|$.

Computing now this order in more detail, we find:
 * $\left|{ \left\langle{a}\right\rangle }\right| \left|{ K }\right| = \left|{ \left\langle{\bar a}\right\rangle }\right| \left({p \left|{ \bar K }\right|}\right)

= p\left({\left|{ \left\langle{\bar a}\right\rangle }\right| \left|{ \bar K }\right|}\right) = p \left|{\bar G}\right| = p^n = \left|{G}\right|$

We conclude that therefore, $G = \langle a \rangle K$.

Using the Internal Direct Product Theorem, we conclude $G = \langle a \rangle \times K$.

The result follows by induction.