Finite direct products are isomorphic up to permutation of the factors

Let G_i be groups for 1 \leq i \leq n and let \pi \in S_n. Prove that the map \varphi_\pi : \times_{i=1}^n G_i \rightarrow \times_{i=1}^n G_{\pi^{-1}(i)} given by (\varphi_\pi(g))_i = g_{\pi^{-1}(i)} for all g = (g_i) is an isomorphism (so that changing the order of the factors in a finite direct product does not change the isomorphism type).

We need to show that \varphi_\pi is a bijective homomorphism.

  1. Homomorphism: Let g = (g_i) and h = (h_i). Then (\varphi_\pi(gh))_i = g_{\pi^{-1}(i)} h_{\pi^{-1}(i)} = (\varphi_\pi(g))_i (\varphi_\pi(h))_i for each i; hence \varphi_\pi(gh) = \varphi_\pi(g) \varphi_\pi(h).
  2. Injective: Let g = (g_i), h = (h_i) \in \times_{i=1}^n G_i such that \varphi_\pi(g) = \varphi_\pi(h). Then for each i, we have g_{\pi^{-1}(i)} = h_{\pi^{-1}(i)}. Because \pi is a permutation of \{1,2,\ldots,n\}, we have g_i = h_i for each i; hence g = h. Thus \varphi_\pi is injective.
  3. Surjective: Let g = (g_{\pi^{-1}(i)}) \in \times_{i=1}^k G_{\pi^{-1}(i)}. It is clear that, with g^\prime = (g_i), we have \varphi_\pi(g^\prime) = g. Thus \varphi_\pi is surjective.
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