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

## Contents

## 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 $K \le G$.

### Corollary

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

Then $G$ can be written as an internal direct product of cyclic groups of prime-power order.

## 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 $\gen a \cap \gen b = \gen e$.

#### Proof of Lemma

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

Hence, $\order c = p$, and $c \notin \gen a$.

From the definition of $b$, we conclude that also $\order b = p$.

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

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

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

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

$\Box$

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

Let $\bar x$ denote the coset $x \gen b$ in $G$.

Assume that $\order {\bar a} < \order a = p^m$.

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

This means:

- $\paren {a \gen b}^{p^{m - 1} } = a^{p^{m - 1} } \gen b = \gen b$

so that:

- $a^{p^{m - 1} } \in \gen a \cap \gen b = \gen e$

This contradicts $\order a = p^m$.

Thus:

- $\order {\bar a} = \order a = p^m$

and therefore $\bar a$ is an element of maximal order in $\bar G$.

As $\order {\bar G} = p^{n - 1}$ 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 preimage of $\bar K$ under the quotient epimorphism from $G$ to $\bar G$:

- $K = \set {x \in G : \bar x \in \bar K}$

As this natural homomorphism is $p$-to-$1$ (that is, every element in the image has $p$ preimages), it follows that:

- $\order K = p \order {\bar K}$

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

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:

- $\order {\gen a K} = \order {\gen a} \order K$

Computing now this order in more detail, we find:

- $\order {\gen a} \order K = \order {\gen {\bar a} } \paren {p \order {\bar K } } = p \paren {\order {\gen {\bar a} } \order {\bar K } } = p \order {\bar G} = p^n = \order G$

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.

$\blacksquare$