Finite NIPC theorem

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This fact is related to: NIPC conjecture
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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.

Related facts


Facts used

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


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.


  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.