# Finite NIPC theorem

This fact is related to: NIPC conjecture
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## Statement

Suppose $G$ is a finite group and $H$ is a normal subgroup of $G$. Then, there exists a finite group $K$ and a surjective homomorphism $\rho:K \to G$ such that both the kernel of $\rho$ and $\rho^{-1}(H)$ are characteristic subgroups of $K$.

## Proof

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

To prove: There exists a group $K$ and a surjective homomorphism $\rho:K \to G$ such that the kernel of $\rho$ and $\rho^{-1}(H)$ are both characteristic in $K$.

Proof:

1. Let $L = G/H$. Suppose $p$ is a prime not dividing the order of $G$. 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 $L = G/H$, a faithful representation of $L$ on $V$ gives a representation of $G$ on $V$ whose kernel is $H$. Let $K$ be the semidirect product $V \rtimes G$ for this action, with $\rho:K \to G$ the quotient map.
3. $V$ (the kernel of $\rho$) is characteristic in $K$: 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. $C_K(V)$ is characteristic in $K$: This follows from the previous step and fact (3).
5. $C_K(V) = V \times H = \rho^{-1}(H)$: Since $V$ is abelian, the quotient group $K/V$ acts on $V$ (fact (4)); 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, $C_K(V) = V \rtimes H$. Since the action is trivial, $C_K(V) = V \times H = \rho^{-1}(H)$.

The last two steps show that $\rho^{-1}(H)$ is characteristic in $K$, while step (3) shows that the kernel of $\rho$ is characteristic in $K$. This completes the proof.