Exercise 14.2.2

Proof. (a) Let $\phi$ the restriction of the sign $\mathrm{sgn}$ to $(S_3\wr S_2)\cap A_6$:

Since $\mathrm{sgn}$ is a morphism, its restriction $\phi$ is also a morphism, and $\phi$ is surjective (onto), because $\phi (e)=1$, and $\phi ((12))=-1$, where $(12)\in S_3\wr S_2$. Moreover the kernel of $\phi$ is $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=(S_3\wr S_2)\cap A_6$.

Therefore $\mathrm{im}(\phi )=\{-1,1\}\simeq (S_3\wr S_2)∕((S_3\wr S_2)\cap A_6)$. This shows that

(b) Let $\tau \in S_n$. Then $\tau$ is in the centralizer of $\sigma =(12)(34)(56)$ if and only if

which is equivalent to

Write $R_1=\{1,2\},R_2=\{3,4\},R_3=\{4,5\}$. Then $R_1,R_2,R_3$ are the three orbits of $\sigma$ acting on $\{1,\dots ,6\}$, the supports of the decomposition of $\sigma$ in disjoint cycles.

Since $\tau$ is a bijection, the 6 values $\tau (1),\tau (2),\tau (3),\tau (4),\tau (5),\tau (6)$ are distinct, so $(\tau (1)\tau (2)),(\tau (3)\tau (4)),(\tau (5)\tau (6))$ are disjoint $2$-cycles.

If $\tau$ is the centralizer of $\sigma$, the equality $(\tau (1)\tau (2))(\tau (3)\tau (4))(\tau (5)\tau (6))=(12)(34)(56)$ shows that $\tau (R_1),\tau (R_2),\tau (R_3)$ are also the three orbits of $\sigma$, so that

that is

which means that there is some permutation ${\tau }^{\prime }$ of $\{1,2,3\}$ such that $\tau (R_i)=R_{{\tau }^{\prime }(i)},i=1,2,3$. In other words, $\sigma$ preserves the blocks $R_1,R_2,R_3$, so that $\sigma \in S_3\wr S_2$.

To prove the converse, it is more convenient to use the other usual representation of $S_3\wr S_2$. Then $\sigma =(e;\mu ,\mu ,\mu )$, where $\mu =(12)\in S_2$. Let $\tau =(\lambda ;{\mu }_1,{\mu }_2,{\mu }_3)$ be any element of $S_3\wr S_2$ (then ${\mu }_i=()$ or ${\mu }_i=\mu )$. Then (14.7) gives

Since $S_2=\{e,\mu \}$ is commutative, $\mu {\mu }_i={\mu }_i\mu ,i=1,2,3$, thus $\mathit{\tau \sigma }=\mathit{\sigma \tau }$.

The centralizer of $(12)(34)(56)$ in $S_n$ is $S_3\wr S_2$. (c) Since the order of $\sigma =(12)(34)(56)$ is $2$, $⟨\sigma ⟩=\{e,\sigma \}\simeq S_2$ and we can write ${\sigma }^{\mathit{𝜀 }},𝜀\in \{0,1\}$ the two elements of $⟨\sigma ⟩$. Let

$\text{∙}$ $\phi$ is a morphism: For all $\tau ,{\tau }^{\prime }\in (S_3\wr S_2)\cap A_6$ and ${\sigma }^{𝜀},{\sigma }^{{𝜀}^{\prime }}\in ⟨\sigma ⟩$, $\sigma {\tau }^{\prime }={\tau }^{\prime }\sigma$ by part (b), thus

$\text{∙}$ $\ker <!--\; FUNCTION\; APPLICATION\; -->\phi$ is trivial: if $\phi (\tau ,{\sigma }^{𝜀})=e$, then $\tau {\sigma }^{𝜀}=e$, so that $\tau ={\sigma }^{-𝜀}\in \{e,\sigma \}$. $\tau =\sigma$ is impossible, since $\tau$ is an even permutation, and $\sigma$ is odd. Therefore $\tau =e$, and ${\sigma }^{𝜀}=e$. Thus $\phi$ is injective (one to one).

$\text{∙}$ Since $|((S_3\wr S_2)\cap A_6)\times ⟨\sigma ⟩|=|S_3\wr S_2|$, $\phi$ is a bijection, thus $\phi$ is a group isomorphism.

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