Let G be a group, and let H = {x/x belongs to G and ax = xa for all a belongs to G}. Show that H is a normal subgroup of G

To show that H is a normal subgroup of G, we need to prove that for every element h in H and every element g in G, the conjugate of h by g, denoted as ghg^(-1), is also in H.

Let’s take an arbitrary element h in H. By the definition of H, we know that for all a in G, ah = ha. Now, we need to show that for any g in G, ghg^(-1) satisfies the condition ax = xa for all a in G.

Consider an arbitrary element a in G. We want to show that (ghg^(-1))a = a(ghg^(-1)).

(ghg^(-1))a = (ghg^(-1))(ag^(-1)g) = g(h(g^(-1)a)g^(-1)) = g(h(g^(-1)a)g^(-1))

Since h is in H, we know that ah = ha for all a in G. Therefore, we can rewrite the expression as:

ghg^(-1)a = g(h(g^(-1)a))g^(-1) = g(ah)g^(-1) = (gag^(-1))(ghg^(-1))

Now, we can see that (ghg^(-1))a = a(ghg^(-1)), which means that ghg^(-1) satisfies the condition ax = xa for all a in G.

Since we’ve shown that for every element h in H and every element g in G, ghg^(-1) is also in H, we can conclude that H is a normal subgroup of G.