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"Let $G_1. G_2$ be groups. Prove that the map

$$ \varphi: G_1 \times G_2 \rightarrow G_2 \times G_1$$ $$ \varphi: (g_1, g_2) \mapsto (g_2, g_1)$$

defines an isomorphism between $G_1 \times G_2$ and $G_2 \times G_1$.

I can prove the homomorphism bit but in the answers, it just says "Obviously $\varphi$ is bijective". Why is this obvious? Is it because it is commutative?

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What did you try ? Did you try proving that it is both injective and surjective ? – Amr Dec 26 '12 at 19:49
I didn't have a clue how to prove it was bijective so I looked at the answers and that's what it said – Kaish Dec 26 '12 at 19:53
@Kaish: to show bijectivity, it also suffices to give an inverse map, namely $G_2\times G_1\to G_1\times G_2$, $(h,g)\mapsto(g,h)$. This is also a homomorphism of course (you showed it the other way round you said; this case is a homomorphism due to symmetry reasons (swap indices)), hence the isomorphy. – InvisiblePanda Dec 26 '12 at 19:56
Just to be sure, there is nothing necessarily commutative here... – copper.hat Dec 26 '12 at 20:19
up vote 3 down vote accepted

If $\varphi(a,b)=(0,0)$ then $a=0$ and $b=0$ so the kernel is trivial and thereby $\varphi$ injective. Now for surjectivity for any $(a,b) \in G_2 \times G_1$ we have that $\varphi(b,a)=(a,b)$.

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Why do we look at the kernel? – Kaish Dec 26 '12 at 19:54
A homomorphism is injective if and only if the kernel is trivial. You can verify this easily by noting that if $\phi(a)=\phi(b)$ and $a \neq b$ then $0=\phi(a)-\phi(b)=\phi(a-b)$. The other direction is even simpler. – JSchlather Dec 26 '12 at 19:55

What is the kernel of $\phi$? Indeed, if $(x,y)\in\ker(\phi)$ where $x\in G_1,y\in G_2$ then $$\phi(x,y)=(y,x)=(e_{G_2},e_{G_1})\Longrightarrow y=e_{G_2},x=e_{G_1}$$ and so the map is a monomorphism (because you showed it is a homomorphism befor). Moreover by taking $(y,x)\in G_2\times G_1$ you have $(x,y)$ in $G_1\times G_2$ and so the map is surjective as well.

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$+\;\;\ddot\smile\;\;+$ – amWhy Mar 2 '13 at 0:48

For injectivity, observe that if $\varphi(g_1,g_2)=\varphi(g_1',g_2')$, then by the defintion of $\varphi$, we get $(g_2,g_1)=(g_2',g_1')$. Hence $g_1=g_1'$ and $g_2=g_2'$ and $\varphi$ is injective.

For surjectively, we ask if for any $(g_2,g_1)\in G_2\times G_1$ we could find some $(a,b)\in G_1\times G_2$ such that $\varphi(a,b)=(g_2,g_1)$. But the answer is clearly affirmative because $\varphi(g_1,g_2)=(g_2,g_1)$.

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The application $\psi$ is defined by: $$\psi: G_2 \times G_1 \rightarrow G_1 \times G_2$$ $$\psi: (g_1,g_2) \mapsto (g_2,g_1)$$ We have: $$\varphi \circ \psi = id_{G_2 \times G_1}$$ $$\psi \circ \varphi = id_{G_1 \times G_2}$$ Therefore $\varphi$ and $\psi$ are bijective and reciprocate.

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