Problem 5.1.7 $G_1 imes G_2 imes \cdots imes G_n \simeq G_{\pi^{-1}(1)} imes G_{\pi^{-1}(2)} imes \cdots imes G_{\pi^{-1}(n)}$

Question

Let $G_1, G_2,\ldots,G_n$ be groups and let $\pi$ be a fixed element of $S_n$. Prove that the map
$$\varphi_\pi : G_1 	imes G_2 	imes \cdots 	imes G_n 	o G_{\pi^{-1}(1)}	imes G_{\pi^{-1}(2)}	imes \cdots 	imes G_{\pi^{-1}(n)}$$
defined by
$$\varphi_{\pi}(g_1,g_2,\ldots,g_n) = (g_{\pi^{-1}(1)},g_{\pi^{-1}(2)},\ldots, g_{\pi^{-1}(n)})$$
is an isomorphism (so that changing the order of the factors in a direct product does not change the isomorphism type).

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $\varphi_\pi$ be the map defined by
$$
\varphi_\pi
\left\{
\begin{array}{ccl}
 G_1 	imes G_2 	imes \cdots 	imes G_n& 	o& G_{\pi^{-1}(1)}	imes G_{\pi^{-1}(2)}	imes \cdots 	imes G_{\pi^{-1}(n)}\
(g_1,g_2,\ldots,g_n) & \mapsto & (g_{\pi^{-1}(1)},g_{\pi^{-1}(2)},\ldots, g_{\pi^{-1}(n)}).
\end{array}
ight.
$$
($\varphi_\pi$ is well defined, since $g_i \in G_i$ for all $i$, so $g_{\pi^{-1}(j)} \in G_{\pi^{-1}(j)}$ for all $j$.)

Consider the map
$$
\psi_\pi
\left\{
\begin{array}{ccl}
 G_{\pi^{-1}(1)}	imes G_{\pi^{-1}(2)}	imes \cdots 	imes G_{\pi^{-1}(n)} & 	o& G_1 	imes G_2 	imes \cdots 	imes G_n  \
(h_1,h_2,\ldots,h_n) & \mapsto & (h_{\pi(1)},h_{\pi(2)},\ldots, h_{\pi(n)}).
\end{array}
ight.
$$
If $h_j = g_{\pi^{-1}(j)}$ and $j = \pi(i)$, then $h_{\pi(i)} = g_{\pi^{-1}(\pi(i))} = g_i$, thus
\begin{align*}
\psi_\pi(\varphi_\pi (g_1,g_2,\ldots,g_n)) &= \psi_\pi(g_{\pi^{-1}(1)},g_{\pi^{-1}(2)},\ldots, g_{\pi^{-1}(n)}) \
&= (g_1,g_2,\ldots,g_n),
\end{align*}
so $\psi_\pi \circ \varphi_\pi = \mathrm{id}$, and similarly $ \varphi_\pi \circ \psi_\pi = \mathrm{id}$. This shows that $\varphi_\pi$ is bijective.

We check that $\varphi_\pi$ is a homomorphism. Pick $g = (g_1,g_2,\ldots,g_n) \in G_1 	imes G_2 	imes \cdots 	imes G_n$ and  $h = (h_1,h_2,\ldots,h_n) \in G_1 	imes G_2 	imes \cdots 	imes G_n$.
Then $k = gh = (k_1,k_2, \ldots, k_n)$, where $k_i = g_i h_i$ for all indices $i$. Then
\begin{align*}
\varphi_\pi(g) \varphi_\pi(h) &= (g_{\pi^{-1}(1)},g_{\pi^{-1}(2)},\ldots, g_{\pi^{-1}(n)}) (h_{\pi^{-1}(1)},h_{\pi^{-1}(2)},\ldots, h_{\pi^{-1}(n)})\
&= (g_{\pi^{-1}(1)}h_{\pi^{-1}(1)}, g_{\pi^{-1}(2)}h_{\pi^{-1}(2)},\ldots, g_{\pi^{-1}(2)}h_{\pi^{-1}(n)}\
&= (k_{\pi^{-1}(1)},k_{\pi^{-1}(2)},\ldots,k_{\pi^{-1}(n)})\
&= \varphi_\pi(k)\
&= \varphi_\pi(gh).
\end{align*}
So $\varphi_\pi$ is an isomorphism, and 
$$G_1 	imes G_2 	imes \cdots 	imes G_n \simeq  G_{\pi^{-1}(1)}	imes G_{\pi^{-1}(2)}	imes \cdots 	imes G_{\pi^{-1}(n)}.$$
\end{proof}

Problem 5.1.7 $G_1 \times G_2 \times \cdots \times G_n \simeq G_{\pi^{-1}(1)}\times G_{\pi^{-1}(2)}\times \cdots \times G_{\pi^{-1}(n)}$

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Problem 5.1.7 $G_1 \times G_2 \times \cdots \times G_n \simeq G_{\pi^{-1}(1)}\times G_{\pi^{-1}(2)}\times \cdots \times G_{\pi^{-1}(n)}$

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