Answer to Question #123757 in Abstract Algebra for Nii Laryea

Question #123757
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
1
Expert's answer
2020-06-29T18:17:08-0400

Given that the map "\\pi : G \u2192 G" defined by "\\pi(g) = g \u2217 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.


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