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## Finite direct products are isomorphic up to permutation of the factors

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.7

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

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 $\varphi_\pi(gh) = \varphi_\pi(g) \varphi_\pi(h)$.

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.

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.