Composition Series of Group of Prime Power Order

Theorem
Let $G$ be a group whose identity is $e$, and whose order is the power of a prime: $\left|{G}\right| = p^n, p \in \mathbb P, n \ge 1$.

Then $G$ has a finite sequence of subgroups:
 * $\left\{{e}\right\} = G_0 \subset G_1 \subset \ldots \subset G_n = G$

such that $\left|{G_k}\right| = p^k$, $G_k \triangleleft G_{k+1}$ and $G_{k+1} / G_k$ is cyclic and of order $p$.

This sequence is called a composition series (or chain of subgroups) for $G$.

Proof
To be proved by induction on $n$.

Let $P_n$ be the proposition for $\left|{G}\right| = p^n$.

Basis for the Induction
$P_1$ is trivially true because:
 * $\left\{{e}\right\} = G_0 \subset G_1 = G$

and a group of order $p$ is cyclic.

Induction Hypothesis
Suppose $P_k$ is true for all groups of order $p^k$ for all $k < n$.

We need to show that this implies $P_{k+1}$ is true.

Induction Step
Let $G$ be a group of order $p^n$.

By Subgroups of Group of Prime Power Order, $G$ has a proper non-trivial normal subgroup.

There will be a finite number of these, so we are free to pick one of maximal order.

We call this $H$, such that $\left|{H}\right| = p^t, t < n$.

We need to show that $t = n - 1$.


 * Suppose $t < n - 1$.

Then $G / H$ is a group of order $p^{n-t} \ge p^2$.

Again by Subgroups of Group of Prime Power Order, $G / H$ has a proper non-trivial normal subgroup, which we will call $N$.

Let $H' = \left\{{g \in G: g H \in N}\right\}$. We now show that $H \triangleleft G$.

Let $g, g' \in H'$.

Then $g H, g' H \in N$.

Since $N < G / H$, $\left({g H}\right) \left({g' H}\right) = g g' H \in N$ and $g g' \in N$.

If $g \in H$, then $g H \in N$.

Since $N < G / H$, $\left({g H}\right)^{-1} = g^{-1} H \in N$, so $g^{-1} \in H'$.

Next:

So clearly $H' / H = N$, therefore:


 * $\dfrac {\left|{H'}\right|} {\left|{H}\right|} = \left[{H' : N}\right] = \left|{N}\right| \ge p$:

So $\left|{H'}\right| \ge p \left|{H}\right|$, contradicting the maximality of $\left|{H}\right|$.

It follows that $t = n - 1$.


 * Finally, we set $G_{n-1} = H$ and use induction to show that $P_{n-1}$ holds.

Since $G / H = G / G_{n-1}$ is a group of order $p$, then it is automatically cyclic.