Exercise 3.4.7 (If $H \unlhd G$, there is a composition series of $G$, one of whose terms is $H$)

Proof. Let $G$ be a nontrivial finite group, and $H⊴G$.

If $H=\{1\}$, or $H=G$, then $H$ is a term of any composition series of $G$, so we may suppose that $H$ is a non trivial proper subgroup.

By the first part of the Jordan-Hölder Theorem (see Ex. 6), $H\ne \{1\}$ has a composition series

where $H_{i+1}∕H_i$ is simple for $i\in [[0,s[[$. Since $H\ne G$, then $G∕H\ne \{\bar{1}\}$, so $G∕H$ has a composition series

where $\overline{K_{i+1}}∕\overline{K_i}$ is simple for $i\in [[0,t[[$. By the Lattice isomorphism Theorem, there exists for every $i\in [[0,t]]$ a unique subgroup $K_i$ of $G$ such that $\overline{K_i}=K_i∕H$, and by part (5) of this Theorem, $K_i⊴K_{i+1}$ ( $0\le i<t$).

Moreover $\{\bar{1}\}=\overline{K_0}=K_0∕H$, so $K_0=H$, and $G∕H=\overline{K_t}=K_t∕H$, so $G=K_t$. This gives

The Third Isomorphism Theorem shows that

therefore $K_{i+1}∕K_i$ is simple for $i\in [[0,t[[$.

If we glue together the chains (1) and (2), we obtain

Put

Since $H_s=H=K_0$, this gives the chain

Moreover, if $0\le i<s$, then $G_{i+1}∕G_i=H_{i+1}∕H_i$ is simple, and if $s\le i<t$, then $0\le i-s<t$ and $G_{i+1}∕G_i=K_{i+1-s}∕K_{i-s}$ is simple.

This shows that (3) is a composition series of $G$, one of whose terms is $H$. □