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I have just encountered the concept of a semi-direct product in group theory and the following exercise is being quite troublesome. If you could give me some help, I will be very grateful. Here we are:

Show that the centralizer of a product of $m$- disjoint $b$-cycles in $S_{mb}$ is isomorphic to a semi-direct product $(\mathbb{Z}/b\mathbb{Z})^m \rtimes S_m$.

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Hint: Let $\sigma\in S_{mb}$ be your product of $m$ disjoint $b$-cycles. As a first step you should be able to show that elements centralizing $\sigma$ permute the orbits of $\langle\sigma\rangle$. – j.p. Nov 15 '12 at 15:39
up vote 4 down vote accepted

We denote by $T$ the “shift” map in $(\mathbb{Z}/b\mathbb{Z})$, defined by $T(x)=x+1$. As a permutation it is a cycle of length $b$. Iterating the action of $T$, we see that the $j$-th iterate $T^j(x)$ is defined for $x$ and $j$ in $\mathbb{Z}/b\mathbb{Z}$ (and in fact, $T^j(x)$ is simply $j+x$ in $(\mathbb{Z}/b\mathbb{Z})$).

Take $m$ copies of $(\mathbb{Z}/b\mathbb{Z})$ : $H_k=(\mathbb{Z}/b\mathbb{Z})$ for $1 \leq k \leq m$, and form the product $H=H_1 \times H_2 \times \ldots \times H_m $. Each $H_k$ has its own private shift $T_k$.

Let us call $\gamma$ the main permutation here. Renaming things, we may assume without loss of generality that $\gamma$ is a permutation of $H_1 \cup H_2 \cup \ldots \cup H_m$, acting on each $H_k$ by “shifting” : $\gamma(x)=T_k(x)=x+1$ for $x\in H_k$.

Denote by $C$ the centralizer of $\gamma$. If $c\in C$, then $c$ commutes with $\gamma$ so $c$ permutes the orbits of $\gamma$ : so we have a permutation $\bar{c}$ of $\lbrace 1,2, \ldots,m \rbrace$ such that $c(H_k)=H_{\bar{c}(k)}$ for every $k$. In particular, if we denote by $0_{H_k}$ the zero element in $H_k$, we have $c(0_{H_k}) \in H_{\bar{c}(k)}$, so that there is an integer $j_k$ such that $c(0_{H_k})=T_{\bar{c}(k)}^{j_k}(0_{\bar{c}(k)})$. The equality $c\gamma=\gamma c$ becomes $cT_k=T_{\bar{c}(k)}c$ on $H_k$, so we deduce

$$ (1) \ \ \ c(x)=T_{\bar{c}(k)}^{j_k+x}(0_{H_{\bar{c}(k)}}) \ (1 \leq k \leq m, x \in H_k) $$

Conversely, the formula (1) uniquely defines a permutation of $H_1 \cup H_2 \cup \ldots \cup H_m$ commuting with $\gamma$. So the map

$$ (2) \ \ \ \phi : C \to (\mathbb{Z}/b\mathbb{Z})^m \rtimes S_m, c \mapsto ((j_1,j_2, \ldots ,j_m),\bar{c}) $$

is a bijection. By construction, it will also be a group homomorphism, qed.

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