1. ## composition series

If $G$ is a finite group and $H$ is a normal subgroup of $G$. Prove that there is a composition series of $G$, one of whose terms is $H$.

Recall that every finite group has a composition series.
Definition: A composition series for $G$ is a series of subgroups of $G$: ${1}$ is a normal subgroup of $N_0$ is a normal subgroup of $N_1$ is a normal subgroup of $...$ is a normal subgroup of $N_k = G$ such that $N_i$ is a normal subgroup of $N_{i+1}$ and $N_{i+1}/N_i$ is simple.

Proof:
$G$ is a finite group, so $G$ has a composition series. And $H$ is a normal subgroup of $G$, by the definition of a composition series, $H$ must be a term of the series.

If $G$ is a finite group and $H$ is a normal subgroup of $G$. Prove that there is a composition series of $G$, one of whose terms is $H$.
Begin by noticing $H\triangleleft G$. If $H$ is a maximal normal subgroup then we can find a composition series $H_0 \triangleleft H_1 ... \triangleleft H_n = H$. Which gives $H_0 \triangleleft H_1 ... \triangleleft H_n = H \triangleleft G$.
If $H$ is not a maximal normal subgroup then there is $H_{(1)}$ so that $H \triangleleft H_{(1)} \triangleleft G$ so that $H_{(1)}\triangleleft G$ is a maximal normal subgroup. Then repeat the argument by finding intermediate normal subgroups between $H$ and $H_{(1)}$. Thus, we can form $H_0 \triangleleft H_1 \triangleleft ... \triangleleft H_n = H \triangleleft ... \triangleleft G$.