Finite Abelian Group is Solvable

Theorem
Let $G$ be a finite abelian group.

Then $G$ is solvable.

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

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * All abelian groups of order $n$ and below are solvable.

Basis for the Induction
$P \left({1}\right)$ is true, as the trivial group is trivially solvable.

From Prime Groups are Simple‎, both $P \left({2}\right)$ and $P \left({3}\right)$ are true.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 3$, then it logically follows that $P \left({k+1}\right)$ is true.

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:
 * $\left|{G}\right| = k+1$

where $\left|{G}\right|$ denotes the order of $G$.

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

From Cauchy's Group Theorem, $G$ 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.