# Composition Series of Group of Prime Power Order

## Theorem

Let $G$ be a group whose identity is $e$, and whose order is a prime power:

- $\order G = p^n, p \in \mathbb P, n \ge 1$

Then $G$ has a composition series:

- $\set e = G_0 \subset G_1 \subset \ldots \subset G_n = G$

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

## Proof

To be proved by induction on $n$.

Let $P_n$ be the proposition for $\order G = p^n$.

### Basis for the Induction

$P_1$ is trivially true because:

- $\set e = G_0 \subset G_1 = G$

From Prime Group is Cyclic, a group whose order is prime 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 Prime Power Group has Non-Trivial Proper Normal Subgroup, $G$ 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 $\order H = p^t, t < n$.

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 Prime Power Group has Non-Trivial Proper Normal Subgroup, $G / H$ has a proper non-trivial normal subgroup, which we will call $N$.

Let $H' = \set {g \in G: g H \in N}$.

We now show that $H \lhd G$.

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

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

Since $N < G / H$:

- $\paren {g H} \paren {g' H} = g g' H \in N$

and so $g g' \in N$.

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

Since $N < G / H$:

- $\paren {g H}^{-1} = g^{-1} H \in N$

and so $g^{-1} \in H'$.

Next:

\(\ds \paren {H'}^a\) | \(=\) | \(\ds \set {g \in G: a g a^{-1} \in H'}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \set {g \in G: a g a^{-1} H \in N}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \set {g \in G: \paren {a H} \paren {g H} \paren {a H}^{-1} \in N}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \set {g \in G: \paren {g H} \in N^{a H} }\) | Definition of Conjugate of Group Subset | |||||||||||

\(\ds \) | \(=\) | \(\ds \set {g \in G: \paren {g H} \in N}\) | as $N$ is normal | |||||||||||

\(\ds \) | \(=\) | \(\ds H'\) |

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

- $\dfrac {\order {H'} } {\order H} = \index {H'} N = \order N \ge p$

So:

- $\order {H'} \ge p \order H$

contradicting the maximality of $\order H$.

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.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 52$