Answer to Question #123757 in Abstract Algebra for Nii Laryea

Question

Let (G, *) be a group. Prove that the map π : G → G defined by π(g) = g * g is a
homomoprhism if and only if G is abelian

Answer 1

Given that the map $\pi : G → G$ defined by $\pi(g) = g ∗ g.$ $\pi(g_1*g_2) = (g_1*g_2)* (g_1*g_2) = g_1*(g_2*g_1)*g_2$ ​ (using associative property of group).

If $(G,*)$ is an abelian group, then $g_2*g_1 = g_1*g_2$

So, $\pi(g_1*g_2)=g_1*(g_2*g_1)*g_2= g_1*(g_1*g_2)*g_2$

$\implies \pi(g_1*g_2) = (g_1*g_1)*(g_2*g_2) = \pi(g_1)*\pi(g_2)$

Hence given mapping is Homomorphism.

Let the given mapping is homomophism, so $\pi(g_1*g_2) = \pi(g_1) * \pi(g_2)$

$\implies g_1*(g_2*g_1)*g_2 = g_1*(g_1*g_2)*g_2$

$\implies g_2*g_1 = g_1*g_2$

Hence, $(G,*)$ is abelian group.

Thus, Given mapping is homomorphism if and only if G is abelian group.