Let $(G, *)$ a group and the two applications $l_g, r_g: G \rightarrow G$ so that $l_g(x) = gx$ and $r_g(x) = xg$. I have already showed that $l_{g_1} \circ l_{g_2} = l_{g_1 g_2}$, $l_g$ is a bijective function but not an homomorphism. Is there a way to prove the same things without doing the hard work? I've seen that $r_g(x) = l_g(g^{-1} x g)$. Does this relation is sufficient to show that $r_{g_1} \circ r_{g_2} = r_{g_1 g_2}$, $r_g$ is a bijective function but not an homomorphism?
EDIT:
I have proved that $c_g: G \rightarrow G$ such that $c_g(x) = g^{-1}xg$ is an isomorphism. Can I use that using the fact that $r_g(x) = l_g(g^{-1} x g) = l_g \circ c_g (x)$?
It is not true that $r_{g_1}r_{g_2}=r_{g_1g_2}$. Rather: $$(r_gr_h)(x)=r_g(r_h(x))=r_g(xh)=(xh)g=x(hg)=r_{hg}(x)$$ and hence $r_{hg}=r_gr_h$. So, $g\mapsto l_g$ is a homomorphism (it's true that $l_g$ is a bijection but not a homomorphism), while $g\mapsto r_g$ is an antihomomorphism.