Talk:Conjugacy class

• C_G is not empty, since (g*g)*g = g*(g*g).
• b * g = g * b → b ^-1 * g = g * b ^-1; thus b ^-1 is in C_G(g)
• (a*b ^-1)*g = a*(b ^-1*g) = a*(g*b ^-1) = (a*g)*b ^-1 = (g*a)*b ^-1 = g*(a*b ^-1) → a*b ^-1 in C_G(g).

thus by elementary group theory, C_G(g) is a subgroup of G.

It is true that C_G(g) is not a normal subgroup of G; but that's not what was stated in the article. Maybe this should be added to the article.User:chas_zzz_brown