1. ## composition series

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

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

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

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