# Finite Abelian Group is Solvable

## Contents

## Theorem

Let $G$ be a finite abelian group.

Then $G$ is solvable.

## Proof

Proof by strong induction on the order of $G$:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

- All abelian groups of order $n$ and below are solvable.

### Basis for the Induction

$\map P 1$ is true, as the trivial group is trivially solvable.

From Prime Group is Simple, both $\map P 2$ and $\map P 3$ hold.

This is the basis for the induction.

### Induction Hypothesis

It remains to be shown that, if $\map P j$ holds, for all $j$ such that $1 \le j \le k$, where $k \ge 3$, then it logically follows that $\map P {k + 1}$ holds.

So this is our induction hypothesis:

- All abelian groups of order $k$ and below are solvable.

Then we need to show:

- All abelian groups of order $k+1$ and below are solvable.

### Induction Step

This is our induction step:

Let $G$ be an abelian group such that:

- $\order G = k + 1$

where $\order G$ denotes the order of $G$.

Then from Positive Integer Greater than 1 has Prime Divisor, there exists some prime $p$ which divides $k + 1$.

From Cauchy's Lemma, $G$ has an element of order $p$.

Since $G$ is abelian, this element generates a subgroup of $G$ which is normal by Subgroup of Abelian Group is Normal.

If $p = n$ and $H = G$ the proof is finished, from Prime Power Group is Solvable.

If $p < n$, then $H$ and $G / H$ (which has order $n / p$) are both solvable from the induction hypothesis.

It follows from Group is Solvable iff Normal Subgroup and Quotient are Solvable that $G$ is solvable.

The result follows by the strong principle of mathematical induction.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Normal and Composition Series: $\S 75$. Solvable Groups