Direct factor implies right-quotient-transitively central factor

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., direct factor) must also satisfy the second subgroup property (i.e., right-quotient-transitively central factor)
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Verbal statement

Any direct factor of a group is a right-quotient-transitively central factor.

Statement with symbols

Suppose G is a group, H \le K \le G, and H is a direct factor of G. Suppose, further, that K/H is a central factor of G/H. Then, K is a central factor of G.


Given: A group G, a direct factor H of G, K contains H and K/H is a central factor of G/H.

To prove: K is a central factor of G.

Proof: Since H is a direct factor of G, there exists a normal complement L to H in G, with HL = G and H \cap L trivial. Let M = K \cap L.

Consider the map \rho:L \to G/H that sends every element of L to its H-coset in G. This map is an isomorphism, since the kernel H \cap L is trivial, and \rho^{-1}(K/H) = M. Thus, HM = K.

  1. Every element of H centralizes every element of L: Since both H and L are normal in G, [H,L] is contained in both, and since they intersect trivially, [H,L] is trivial, so every element of H centralizes every element of L.
  2. MC_L(M) = L: Since \rho is an isomorphism, and \rho(M) is a central factor of G/H, M is a central factor of L.
  3. C_L(M) centralizes K, i.e., C_L(M) \le C_G(K): By step (1), C_L(M) centralizes H, since L centralizes H. It also centralizes M by definition. Thus, it centralizes HM = K.
  4. KC_L(M) = G: We have KC_L(M) = HMC_L(M) = HL = G.
  5. KC_G(K) = G: This follows from the previous two steps.

This completes the proof.