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 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 $$\left|{G}\right| = p^n$$.


 * $$P_1$$ is trivially true because $$\left\{{e}\right\} = G_0 \subset G_1 = G$$, and a group of order $$p$$ is cyclic Group of Prime Order Cyclic.


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

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, so we do that and call it $$H$$, such that $$\left|{H}\right| = p^t, t < n$$.

What we want to do is 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:

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

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.