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...


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...

  • 5
    $\begingroup$ The group must be abelian? $\endgroup$ – Just do it Oct 22 '18 at 12:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.