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I am studying a text on permutation groups, which has the following example in a section on regular normal subgroups:

If $Z(N)=1$, then $N \cong \mathrm{Inn}(N)$, the group of inner automorphisms of $G$, and the semidirect product $N \rtimes \mathrm{Inn}(N)$ is the diagonal group $N^*=N \times N$ described earlier.

My question is on why the group $N \rtimes \mathrm{Inn}(N)$ "is" the group $N \times N$? What would be the group isomorphism? I presume it is not necessary that $N$ be regular and normal since this is not stated explicitly in this example, and I'm not sure why a $G$ enters the picture (is this a typo - should this be an $N$ instead?).

Let me recall a couple facts described earlier in the text:

(a) If $G$ is a group, then $G^* := G \times G$ acts on $G$ by the rule $\mu(x,(g,h)):=g^{-1} xh$. Thus, $G^*$ is the product action that contains the left and right regular actions as normal subgroups.

(b) If $N$ is a regular normal subgroup of $G$, then $G=N \rtimes G_1$; and conversely, if $N$ is any group and $H \le \mathrm{Aut}(N)$, then $N \rtimes H$ acts as a permutation group on $N$, with $N$ as a regular normal subgroup and $H$ as the stabilizer of the identity; the element $hn$ of $N \rtimes H$ acts on $N$ by the rule $\mu(x,hn)=x^h n$. [The group operation in $N \rtimes H$ is not specified in the text, but we do get an action if the binary operation is assumed to be $(n_1,h_1)(n_2,h_2)=(n_1^{h_2} n_2, h_1 h_2)$.]

Since $N/Z(N) \cong \mathrm{Inn}(N)$, the assumption $Z(N)=1$ implies $N \cong \mathrm{Inn}(N)$. To get an isomorphism from $N^*:= N \times N$ to $N \rtimes \mathrm{Inn}(N)$, I tried the map $(g,h) \mapsto (g, c(h))$, where $c(h)$ denotes conjugation by $h$. But this map does not seem to be a homomorphism: $\phi((g_1,h_1)(g_2,h_2))= \phi((g_1 g_2, h_1 h_2)) = (g_1 g_2, c(h_1 h_2))$, but $\phi((g_1,h_1)) \phi((g_2,h_2)) = (g_1, c(h_1))(g_2,c(h_2)) = (g_1 ^{c(h_2)} g_2, c(h_1 h_2))$.

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In the semidirect product $N \rtimes \operatorname{Inn} N$, the normal subgroup $N \times 1$ is centralized by the subgroup $\{ (h^{-1},h) \mid h \in N \}$, which is isomorphic to $N$. So this subgroup is equal to the second factor $1 \times N$ in the direct product $1 \times N$.

The isomorphism $N \times N \to N \rtimes \operatorname{Inn} N$ is given by $(g,h) \to (h^{-1}g,h)$.

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If $N \trianglelefteq H$, then $H$ acts on $N$ by conjugation. If this action induces only inner automorphisms (ie. for all $h \in H$, the map $n \mapsto n^h$ is an inner automorphism of $N$), it follows that $H = N C_H(N)$. If also $Z(N) = 1$, then $H = N \times C_H(N)$.

Thus in your case of $H = N \rtimes \operatorname{Inn}(N)$, it follows that $H = N \times C_H(N)$. Here $C_H(N) \cong H/N \cong \operatorname{Inn}(N) \cong N$, so $H \cong N \times N$.

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