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 $\gen a \times K$ for some $K \le G$.

Proof
Suppose $\order G = p^n$ with $p$ a prime.

We proceed by induction on $n$.

Basis for the induction
For $n=1$, we have $G = \gen a \times \gen e$, by Prime Group 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 $\order x \divides \order G$ by Order of Element Divides Order of Finite Group.

Thus we have $\order x = p^i$, for some $i \ge 0$.

By assumption, $i \le m$.

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

Now assume $G \ne \gen a$, for the remaining case is trivial.

Choose $b \in G$ such that $\order c < \order b$ implies $c \in \gen a$, but $b \notin \gen a$.

Lemma
We have $\gen a \cap \gen b = \gen e$.

 Since $\order {b^p} = \dfrac {\order b} p < \order b$, from the definition of $b$ we conclude $b^p \in \gen a$.

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

Note that $e = b^{p^m} = \paren {b^p}^{p^{m - 1} } = \paren {a^i}^{p^{m - 1} }$, so $\order {a^i} \le p^{m - 1}$.

Hence $a^i$ is not a generator of $\gen a$.

Therefore $\gcd \set {p^m, i} \ne 1$.

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 $\gen a$, 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 $c \notin \gen a$.

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

It follows from Prime Group is Cyclic that $\gen b$ is generated by all its non-identity elements.

Therefore, by Intersection of Subgroups, $\gen a \cap \gen b$ is either $\gen b$ or $\gen e$.

However, the first case contradicts $b \notin \gen a$.

We conclude that $\gen a \cap \gen b = \gen e$.

Now consider the factor group $\bar G = G / \gen b$.

Let $\bar x$ denote the coset $x \gen b$ 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 $\left({a \gen b}\right)^{p^{m-1}} = a^{p^{m-1}} \gen b = \gen b$, so that $a^{p^{m-1}} \in \gen a \cap \gen b = \gen e$.

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 $\gen {\bar a} \times \bar K$ for some subgroup $\bar K$ of $\bar G$.

Let $K$ be the pullback of $\bar K$ under the quotient epimorphism 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 $x \in \gen a \cap K$, then we have $\bar x \in \gen {\bar a} \cap \bar K = \gen {\bar e} = \gen {\bar b}$.

It follows that $x \in \gen a \cap \gen b = \gen e$ and therefore, $\gen a \cap K = \gen e$.

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

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

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

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

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

The result follows by induction.