Finite NPC theorem

Statement

Suppose $K$ is a finite group and $H$ is a normal subgroup of $K$. Then, there exists a finite group $G$ containing $K$ such that $H$ is a characteristic subgroup of $G$.

Facts used

1. Cayley's theorem
2. Normal Hall implies characteristic
3. Characteristicity is centralizer-closed
4. Quotient group acts on abelian normal subgroup
5. Characteristicity is transitive

Proof

Given: A finite group $K$, a normal subgroup $H$ of $K$.

To prove: There exists a group $G$ containing $K$ such that $H$ is characteristic in $G$.

Proof:

1. Let $L=K/H$. Suppose $p$ is a prime not dividing the order of $K$. By fact (1), $L$ is a subgroup of the symmetric group $\operatorname {Sym} (L)$, which in turn can be embedded in the general linear group $GL(n,p)$ where $n=|L|$. Thus, $L$ has a faithful representation on a vector space $V$ of dimension $n$ over the prime field of order $p$.
2. Since $L=K/H$, a faithful representation of $L$ on $V$ gives a representation of $K$ on $V$ whose kernel is $H$. Let $G$ be the semidirect product $V\rtimes K$ for this action.
3. $V$ is characteristic in $G$: In fact, $V$ is a normal $p$-Sylow subgroup, and hence is characteristic (fact (2)) (it can be defined as the set of all elements whose order is a power of $p$).
4. $C_{G}(V)$ is characteristic in $G$: This follows from the previous step and fact (3).
5. $C_{G}(V)=V\times H$: Since $V$ is abelian, the quotient group $G/V$ acts on $V$; in particular, any two elements in the same coset of $V$ have the same action by conjugation on $V$. Thus, the centralizer of $V$ comprises those cosets of $V$ for which the corresponding element of $G$ fixes $V$. This is precisely the cosets of elements of $H$. Thus, $C_{G}(V)=V\rtimes H$. Since the action is trivial, $C_{G}(V)=V\times H$.
6. $H$ is characteristic in $V\times H$: $H$ is a normal subgroup of $V\times H$, on account of being a direct factor. Further, it is a normal $p'$-Hall subgroup, so by fact (2), it is characteristic in $V\times H$.
7. $H$ is characteristic in $G$: By steps (4) and (5), $V\times H$ is characteristic in $G$, and by step (6), $H$ is characteristic in $V\times H$. Thus, by fact (5), $H$ is characteristic in $G$.