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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Statement

Verbal statement

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

Statement with symbols

Suppose G is a group, HKG, 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.

Proof

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 HL trivial. Let M=KL.

Consider the map ρ:LG/H that sends every element of L to its H-coset in G. This map is an isomorphism, since the kernel HL is trivial, and ρ1(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. MCL(M)=L: Since ρ is an isomorphism, and ρ(M) is a central factor of G/H, M is a central factor of L.
  3. CL(M) centralizes K, i.e., CL(M)CG(K): By step (1), CL(M) centralizes H, since L centralizes H. It also centralizes M by definition. Thus, it centralizes HM=K.
  4. KCL(M)=G: We have KCL(M)=HMCL(M)=HL=G.
  5. KCG(K)=G: This follows from the previous two steps.

This completes the proof.