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Symmetric group acts on the n-fold direct product of a group by permuting the factors

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.8

Solution: First we show that this mapping $\psi : \pi \mapsto \varphi_\pi$ is a homomorphism. Let $g = (g_i) \in G$.

We have \begin{align*}[(\varphi_{\pi_1} \circ \varphi_{\pi_2})(g)]_i =&\ [\varphi_{\pi_1}(\varphi_{\pi_2}(g))]_i \\=&\ [\varphi_{\pi_2}(g)]_{\pi_1^{-1}(i)}\\ =&\ g_{\pi_2^{-1}(\pi_1^{-1}(i))} \\=&\ g_{(\pi_2^{-1} \circ \pi_1^{-1})(i)}\\ =&\ g_{(\pi_1 \circ \pi_2)^{-1}(i)}\\ =&\ [\varphi_{\pi_1 \circ \pi_2}(g)]_i. \end{align*}Thus $$\varphi_{\pi_1} \circ \varphi_{\pi_2} = \varphi_{\pi_1\pi_2},$$ hence this mapping is a homomorphism.

Now suppose the $G_i$ are nontrivial and let $\pi$ be in the kernel of $\psi$. Then for all $g \in G$, $$g_i = (\varphi_\pi(g))_i = g_{\pi^{-1}}(i).$$ Since the $G_i $ are nontrivial, we may choose $g$ such that any given pair of entries consists of distinct elements of $G_i$. Thus $\pi(i) = i$ for all $i$, and in fact $\pi = 1$. Thus the kernel of $\psi$ is trivial, hence $\psi$ is injective.

Note that the nontrivial condition on $G_i$ is necessary because otherwise, $G \cong 1$ and $\mathsf{Aut}(G) \cong 1$.


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