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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?

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You already proved that $\phi$ is a homomorphism. Then the easiest way to show that this map is bijective is to determine explicitly the inverse homomorphism. – Curufin Dec 12 '12 at 15:52
Whats the inverse homomorphism? Basically saying that $\varphi(a) \varphi(b) = \varphi(ab)$ and $\varphi(1_H) = 1_G$? – Kaish Dec 12 '12 at 16:05
Think about the inverse of the permutation. – Gregor Bruns Dec 12 '12 at 16:08
By definition, the permutation is a bijection to itself. And as we are mapping to permutations, thus the original group must be a permutation and so there is a bijection? – Kaish Dec 12 '12 at 16:51
up vote 0 down vote accepted

$$\text{If}\,\,\,\phi:G_1\times\ldots\times G_k\to G_{\sigma(1)}\times\ldots\times G_{\sigma(k)}\,\,\,\text{is defined by}\,\,\phi(g_1,...,g_k):=(g_{\sigma(1)},...,g_{\sigma(k)})\,\,, $$

$$\text{define then }\,,\pi: G_{\sigma(1)}\times\ldots\times G_{\sigma(k)}\to G_1\times\ldots\times G_k\,\,\,\text{by:}$$


which, in fact, is just putting $\,\pi(x_1,...,x_k):=\left(x_{\sigma^{-1}(1)},...,x_{\sigma^{-1}(k)}\right)\,$.

Now just show that $\,\phi\circ\pi=\pi\circ\phi=Id.\,$ ,and, thus, $\,\phi\,$ is a bijection (just as you showed $\,\phi\,$ is a homomorphism you can show the same for $\,\pi\,$) and thus an isomorphism.

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