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Determine all groups $G$ for which the map $\phi:G\rightarrow G$ defined by $\phi(g) = g^2$ is a homomorphism.

I have problems understanding the task, what is it: determine all such groups? Find a property that they all have? I checked the axioms of homomorphism, like nothing supernatural of them goes for the structure of the image.

$\phi(g_1 g_2) = (g_1 g_2)^2 = g_1^2 g_2^2 = \phi(g_1)\phi(g_2)$

$\phi(1_G) = 1_G^2 = 1_G$

$\phi(g_1^{-1})=(g_1^{-1})^2 = g_1^{-2}$

$\phi(g_1 g_2)^{-1} = (g_1 g_2)^{-2} = g_2^{-2} g_1^{-2} = \phi(g_2^{-1})\phi(g_1^{-1})$

$\phi(g^0) = (g^0)^2 = g^0 = 1_G$

$\phi(g g^{-1}) = (g g^{-1})^2 = g^2 g^{-2} = 1_G$

I have no idea about what I should check...

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Hint: you've written $(g_1g_2)^2 = g_1^2g_2^2$. But $(g_1g_2)^2 = g_1g_2g_1g_2$ doesn't equal $g_1^2g_2^2$ in general. Think about what is required for this to be true...

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    $\begingroup$ The group must be abelian? $\endgroup$ – Just do it Oct 22 '18 at 12:29

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