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Let $G = A \times B$. Suppose that $H \leq G$ such that $N_G(H) = N_A(\pi_A(H)) \times N_B(\pi_B(H))$ where $\pi_A$ and $\pi_B$ are the respective projection homomorphisms

For simplicity and convenience, we can identify $G$ as being an internal direct product of $A$ and $B$

Proposition 1: $N \unlhd A \times B$ if and only if $[N, X] = [\pi_X(N), X] \leq N \cap X$ for $X = A, B$

Definition For $H \leq A \times B$, we define $C_X = \{x \in X \,|\, [\,x, \pi_X(H)\,] \leq X \cap H \}$ for $X = A, B$.

Proposition 2: If $H \leq A \times B$ then $N_G(H) \cap X = C_X$ for $X =A, B$.

Question: I need to show that if $N_G(H) = N_A(\pi_A(H)) \times N_B(\pi_B(H))$ then $N_X(\pi_X(H)) \leq C_X$ for $X = A, B$.

We prove this for $X =A$. Since $H \unlhd N_A(\pi_A(H)) \times N_B(\pi_B(H)) = N_G(H)$, we can apply Proposition 1, to obtain that $[H, N_A(\pi_A(H))] \leq H \cap N_A(\pi_A(H)$.

Let $x \in N_A( \pi_A(H)$. To show that $x\in C_A$ i.e $ [x, \pi_A(H)] \leq H \cap A$

Now $\pi_A(H) \unlhd N_A( \pi_A(H)$, hence $ [x, \pi_A(H)] \ \subseteq \pi_A(H)$. Also $H \cap N_A(\pi_A(H)) \leq H \cap A$

If I can show that $ \pi_A(H) \leq [H, N_A(\pi_A(H))]$, I would be done but I'm not sure how to proceed. If anyone can be of assistance, it will be greatly appreciated.

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    $\begingroup$ Doesn't this follow immediately from Proposition 2? $\endgroup$
    – Derek Holt
    Apr 3, 2017 at 9:56
  • $\begingroup$ Okay, we can identify $N_A( \pi_A(H))$ as being a subgroup of $N_G(H)$ and since $N_A(\pi_A(H)) \leq A$. Then $N_A( \pi_A(H)) \leq N_G(H) \cap A = C_A$. Is this correct? The journal paper that I am using proposed the proof that I tried to carry out above. I'm not sure why though $\endgroup$
    – R Maharaj
    Apr 3, 2017 at 12:50

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