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:
- $\order {G_k} = p^k$
where:
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.
We have that $G / H = G / G_{n - 1}$ is a group of order $p$
From Prime Group is Cyclic, it follows that $G / H$ is cyclic.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 52$