Let G be a group and a∈G. Prove that <a>C(a) ≤ G where <a>C(a) = {bc|b∈<a>, c∈C(a)}?

Question

Need to prove that <a>C(a) ≤ G where <a>C(a) = {bc|b∈<a>, c∈C(a)}

Answer 1

Let x and y be elements of <a>C(a)

Then x = Bx*Cx and y = By*Cy for Bx,By in <a> and Cx,Cy in C(a)

y_inverse = (Cy_inverse * By_inverse) <--- subgroups <a> and C(a) contain the respective inverses

x*y_inverse = Bx*Cx * (Cy_inverse * By_inverse) <--- substitution

= Bx * (Cx * Cy_inverse) * By_inverse <--- associative property holds for groups

= Bx * C_n * By_inverse <--- closure property holds for group where C_n = Cx*Cy_inverse

= Bx * By_inverse * C_n <--- commutative property holds for C(a)

= (Bx * By_inverse) * C_n <--- associative property again

= B_n * C_n <--- closure property again

So x*y_inverse is of the proper form where B_n and C_n are in their respective subgroups.

The one-step proof is complete.