Solving Simple Group Homomorphism Problem: Proving phi(G) is Subgroup of N

[A.Magnus]

If $A$ isn't a group then $A^{-1}$ might be a different set. If you want one more question to test your understanding, then if $A$ is a group, what is $AA$?

I think $AA = A^2 = A$, because all elements are closed by operation. Correct?

 

[Dick]

I think $AA = A^2 = A$, because all elements are closed by operation. Correct?

Correct. Another way to see it is that $e$ is an element of $A$. So $A=eA \subset AA$, and, of course $AA$ can't contain any elements that aren't in $A$ since group multiplication is a closed operation. As you say. I think you've got it.