Length of Subgroup Plus Length of Quotient Group

Theorem
Let $G$ be a finite group.

Let $H$ be a normal subgroup of $G$.

Then:
 * $\map l G = \map l H + \map l {G / H}$

where:
 * $\map l G$ denotes the length of $G$
 * $G / H$ denotes the quotient group of $G$ by $H$.

Proof
Let $\HH$ be a composition series of $H$:


 * $\set e = H_0 \subset H_1 \subset \dots \subset H_m = H$,

We can construct a composition series $\GG$ for $G$ which contains $H$:


 * $\set e = H_0 \subset H_1 \subset \dots \subset H_m = H = G_0 \subset G_1 \subset \dots \subset G_n = G$

where, for $k = 0, \dots, n - 1$:


 * $G_k \vartriangleleft G_{k + 1}$

and there exists no normal subgroup $G'$ of $G$ such that:


 * $G_k \subset G' \subset G_{k + 1}$.

Now consider the sequence $\SS$:


 * $H = G_0 \subset G_1 \subset \dots \subset G_n = G$

We can construct a normal series $\GG_1$ for $G / H$ from $\SS$:


 * $H = G_0 / H \subset G_1 / H \subset \dots \subset G_n / H = G / H$

$\GG_1$ is indeed a normal series for $G / H$ because:


 * $G_k \vartriangleleft G_{k + 1}$ by construction
 * $H \vartriangleleft G_k$
 * $\paren {G_{k + 1} / H} / \paren {G_k / H} \cong G_{k + 1} / G_k$ by Third Isomorphism Theorem.

there exists a normal subgroup $G' / H$ of $G / H$ such that:


 * $G_k / H \subset G' / H \subset G_{k + 1} / H$

for $k = 0, \dots, n - 1$.

Then the Third Isomorphism Theorem implies:
 * $\paren {G' / H} / \paren {G_k / H} \cong G' / G_k$

Then $G_k$ is a proper normal subgroup of $G'$, contradicting the construction of $\GG$.

So no such $G' / H$ exists.

Hence, $\GG_1$ has no proper refinements and is a composition series.

Since the isomorphism $\paren {G_{k + 1} / H} / \paren {G_k / H} \cong G_{k + 1} / G_k$ holds for all $G_k, G_{k + 1}$:


 * the length of $\SS$ is equal to $\map l {\GG_1} = \map l {G / H}$.

The result follows from the construction of $\GG$.