In a semidirect product, the centralizer of the normal factor in the nonnormal factor is the kernel of the defining homomorphism

Let H and K be groups, let \varphi : K \rightarrow \mathsf{Aut}(H) be a homomorphism, and let G = H \rtimes_\varphi K. Identify H = H \times 1 and K = 1 \times K as subgroups of G.

Prove that C_K(H) = \mathsf{ker}\ \varphi. (Recall that C_K(H) = C_G(H) \cap K.)

(\subseteq) Let (1,k) \in C_K(H). By definition, if (h,1) \in H \times 1, then (h,1) \cdot (1,k) = (1,k) \cdot (h,1). By the definition of multiplication in a semidirect product, we have (h \cdot \varphi(1)(1), k) = (1 \cdot \varphi(k)(h), k), so that (h,k) = (\varphi(k)(h),k). Comparing first coordinates, h = \varphi(k)(h). Since h \in H is arbitrary, we have \varphi(k) = \mathsf{id}_H, so that k \in \mathsf{ker}\ \varphi.

(\supseteq) Suppose (1,k) \in \mathsf{ker}\ \varphi \leq K. Then \varphi(k) = \mathsf{id}_H. Now let (h,1) \in H \times 1 be arbitrary. Then (1,k) \cdot (h,1) = (\varphi(k)(h), k) = (h,k) = (h \cdot \varphi(1)(1), k) = (h,1) \cdot (1,k). Thus (1,k) \in C_K(H).

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  • Gobi Ree  On January 6, 2012 at 2:56 am

    If we identify H and K with their isomorphic copies in G, it will be much simpler: k \in C_K(H) iff khk^{-1}=\varphi(k)(h)=h, \forall h \in H iff k \in \mathsf{ker} \varphi.

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