Theorem The solvable simple groups have prime order.
proof: Let $G$ be solvable then $G' = [G,G] \lhd G$ by the optimal-series theory, if $G$ is simple then $G' = 1$ so $G$ is abelian and the result follows from...
Lemma The abelian simple groups are the cyclic groups of prime order.
proof: By Lagrange's theorem if a group has prime order it has no nontrivial subgroups. Every subgroup of an abelian group is normal, so we need to classify the abelian groups that have no nontrivial subgroups. The group must be cyclic since we need every element to be a generator (except 1) so we see it's at least $C_{p^r}$ but $C_{p^{r-1}}$ (by raising everything to the $p$th power) is a subgroup of that and that's only nontrivial when $r=1$.
Theorem $G$ is solvable iff every composition factor has prime power order.
proof: If $G$ is solvable then any factor of its series is solvable by basic-properties, but such factors are also simple because we have a composition series (one which admits no refinement), so we are done by the above theorem.
Definition A subgroup $N$ of $G$ is a minimal normal subgroup if $N \unlhd G$ and for any $M \unlhd G$ with $M \le N$ either $1 = M$ or $M = N$.
Definition A group is an elementary abelian $p$-group if it is a direct product of copies of the cyclic group $C_p$.
Proposition A minimal normal $N$ subgroup of a solvable group is elementary abelian.
proof: Since it's a subgroup of a solvable group $N$ is solvable too, so it's derived series terminates thus $N' = [N,N] < N$, but $N' \text{ char } N$ and $N \lhd G$ so $N' \lhd G$ implies $N' = 1$. Therefore $N$ is abelian so by the structure theorem it's a product of cylic groups, let $p \mid |N|$ and consider the normal subgroup $M$ of order $p$ elements: $M\text{ char }N$ so $M=N$, every factor is $C_p$.
No comments:
Post a Comment