Isomorphism of subgroup of $S_n$.

group-isomorphismgroup-theorypermutationssymmetric-groups

Let $n,k \in \mathbb{N}$ with $1<k<n$. We define a subgroup $H \leq S_n$ as:
$$H=\{\sigma \in S_n | \; \sigma(\{1,…,k\}) \subseteq \{1,…,k\}\}$$
I have to show that $H$ is isomorphic to $S_k\times S_{n-k}$.

My idea is that for all permutations in $H$, we can divide it in two parts, one of them only permutates elements from $\{1,…,k\}$ and the other one permutates only elements from $\{k+1,…,n\}$, so I could write $\sigma \in H$ as $\sigma = \rho\tau, \rho \in S_k$ and $\tau \in S_{n-k}$.

With that, an isomorphism should be easy to define, but I don't really know how can I formalize that idea (or if it's really right). Any hints? Thanks.

Best Answer

Deploying my hint in the comment, let's define $I:=\{1,\dots,k\}$ and $J:=\{k+1,\dots,n\}$:

  1. Injectivity: for every $\sigma,\tau\in S_n$, \begin{alignat}{2} &\varphi(\sigma)=\varphi(\tau) &&\Longrightarrow \\ &(\sigma_{|I}=\tau_{|I}) \wedge (f\sigma_{|J}f^{-1}=f\tau_{|J}f^{-1}) &&\Longrightarrow \\ &(\sigma_{|I}=\tau_{|I}) \wedge (\sigma_{|J}=\tau_{|J}) &&\Longrightarrow \\ &\sigma=\tau \end{alignat}
  2. Surjectivity: for every $(\psi_1,\psi_2)\in S_k\times S_{n-k}$, define $\sigma\in S_n$ by $\sigma_{|I}:=\psi_1$ and $\sigma_{|J}:=f^{-1}\psi_2f$; then $(\psi_1,\psi_2)=\varphi(\sigma)$.
  3. Preservation of the operation (composition): for every $\sigma,\tau\in H$ and $i\in I$, \begin{alignat}{2} (\sigma\tau)_{|I}(i) &\stackrel{i\in I}{=} &&(\sigma\tau)(i) \\ &\stackrel{\sigma,\tau\in S_n}{=} &&\sigma(\tau(i)) \\ &\stackrel{\tau\in H}{=} &&\sigma_{|I}(\tau(i)) \\ &\stackrel{i\in I}{=} &&\sigma_{|I}(\tau_{|I}(i)) \\ &\stackrel{\sigma_{|I},\tau_{|I}\in S_k}{=} &&(\sigma_{|I}\tau_{|I})(i) \\ \end{alignat} whence $(\sigma\tau)_{|I}=\sigma_{|I}\tau_{|I}$. Likewise, $(\sigma\tau)_{|J}=\sigma_{|J}\tau_{|J}$, because $\lambda\in H\iff \lambda\in K:=\{\gamma\in S_n\mid \gamma(J)\subseteq J\}$. Therefore: \begin{alignat}{1} \varphi(\sigma\tau) &= ((\sigma\tau)_{|I}, f(\sigma\tau)_{|J}f^{-1}) \\ &= (\sigma_{|I}\tau_{|I}, f\sigma_{|J}\tau_{|J}f^{-1}) \\ &= (\sigma_{|I}\tau_{|I}, (f\sigma_{|J}f^{-1})(f\tau_{|J}f^{-1})) \\ &= (\sigma_{|I}, f\sigma_{|J}f^{-1})(\tau_{|I}, f\tau_{|J}f^{-1}) \\ &= \varphi(\sigma)\varphi(\tau) \end{alignat}

and $H \cong S_k\times S_{n-k}$.

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