Problem

If \(G\) is a finite group and \(H \le G\) prove that there is a composition series of \(G\), one of whose terms is \(H\).

Solution

If \(H = 1\), then \(H = 1\) is part of a composition series \(1 = H \trianglelefteq … \trianglelefteq G\) of \(G\). For the rest, assume \(H \ne 1\).

Since \(H\) is finite, by the first part of Jordan-Hölder Theorem, \(H\) has a composition series, say,

$$ 1 = H_0 \trianglelefteq … \trianglelefteq H_k = H $$

for some positive integer \(k\).

\(G/H\) is also finite, and thus it has a composition series, say,

$$ 1 = H/H \trianglelefteq H_{k+1}/H \trianglelefteq … \trianglelefteq H_n/H = G/H $$

for some positive integer \(n\).

Consider each pair \((H_i/H, H_{i+1}/H)\) of the composition series of \(G/H\) where \(k < i \le n-1\). Since \(H_i/H \trianglelefteq H_{i+1}/H\), by the Fourth Isomorphism Theorem, \(H_i \trianglelefteq H_{i+1}\).

Note

• \(H\) and \(H_i\) are both normal subgroups of \(H_{i+1}\),
• \(H \le H_i\), and
• \((H_{i+1}/H)/(H_i/H)\) is simple.

Applying this to the Third Isomorphism Theorem, \(H_{i+1}/H_i\) is simple.

Therefore, combining the two series above,

$$ 1 = H_0 \trianglelefteq … \trianglelefteq H_{k-1} \trianglelefteq H \trianglelefteq H_{k+1} \trianglelefteq … \trianglelefteq H_n = G $$

is a composition series of \(G\). Notice it contains \(H\).

$$\tag*{$\blacksquare$}$$