Prove that the map:
$ \varphi: G_1 \times \cdots \times G_k \rightarrow G_{\sigma(1)} \times \cdots \times G_{\sigma(k)} $ $ \varphi: \hspace{0.2cm} (g_1, \cdots , g_k) \hspace{0.2cm} \mapsto \hspace{0.5cm}(g_{\sigma(1)}, \cdots ,g_{\sigma(k)}) $
defines an isomorphism of the groups.
This is my proof:
1) I don't know how to show its bijective. Can you say that it is obvious as the groups map to symmetric perumatations? I don't know why that would show bijection though.
2) $\varphi(ab) = \varphi(a) \varphi(b)$:
$\varphi(g_1 , \cdots, g_k) = \varphi(g_1 \cdots g_k) = g_{\sigma(1)} \cdots g_{\sigma(k)} = \varphi(g_{\sigma(1)}) \cdots \varphi (g_{\sigma(k)}) $
3) $\varphi(1_G) = 1_H$:
$\varphi(1_{G_1 \times \cdots \times G_k}) =\varphi(1_{G_1} \times \cdots \times 1_{G_k}) = (1_{G_{\sigma(1)} }, \cdots, 1_{G_{\sigma(1)}}) = 1_{G_{\sigma(1)} \times \cdots \times G_{\sigma(k)}}$
Is this correct?
EDIT: Does commutativity show bijection?