How to show that the commutator subgroup is a normal subgroup It is suggested as an exercise in Serge Lang's book "Algebra" to show that the commutator subgroup $G^c$ of a group $G$ is a normal subgroup. 
I'd like to do that but I am afraid I need help, 
I think the first thing I need to figure out is how a general element in the commutator subgroup looks like, so that I can check that the defining condition for normality is satisfied.
That is, supposing for a moment that a general element in $G^c$ is denoted by $g$, I need to show that $aga^{-1} \in G^c,$ for all $a \in G$.
But here I get stuck, first because I am unsure how to write a general element in $G^c$ - a simple product in $G^c$ is of the form $xyx^{-1}y^{-1}aba^{-1}b^{-1}$ where $a,b \in G$. I cannot see a way to simplify this - I am sure there is one, but somehow I am blind today.
The second thing then is, even if one tries out the conjugation of a simple element like $xyx^{-1}y^{-1}$ in $G^c$, again not simplification offers itself easily I think .. what am I missing ?
An alternative would be to find a homomorphism of $G$ whose kernel is precisely $G^c$ - here I tried to think of this as a map $G \times G \to G$ but whatever I cook up is not a homomorphism.
Thanks for your hints !! 
 A: Denote the commutator of $a$ and $b$ by $a^{-1}b^{-1}ab = [a,b]$.
If $u$ is an element from the commutator subgroup, then $g^{-1}ug = u(u^{-1}g^{-1}ug) = u[u, g]$ .

Another approach: the commutator subgroup is defined to be the subgroup generated by the commutators, so every element of the commutator subgroup is of the form $$[a_1, b_1][a_2,b_2]\ldots[a_n, b_n].$$ It is enough to show that $g^{-1}[a,b]g$ is always in the commutator subgroup, because then
$$g^{-1}[a_1, b_1][a_2,b_2]\ldots[a_n, b_n]g = (g^{-1}[a_1, b_1]g)(g^{-1}[a_2,b_2]g)(g^{-1}\ldots g)(g^{-1}[a_n, b_n]g)$$
is a product of elements from the commutator subgroup. When $\phi$ is any homomorphism, we have $\phi([a,b]) = [\phi(a), \phi(b)]$. Since for any $g \in G$ the map $\phi$ defined by $\phi(x) = g^{-1}xg$ is a homomorphism, the result follows. 
A: If $c \in G'$ and $g \in G$, then also $[g, c] = gc g^{-1} c^{-1} \in G'$. Because G' is closed under products, we would also have $(gc g^{-1} c^{-1})c \in G'$. But $(gc g^{-1} c^{-1})c = gcg^{-1} \in G'$, so by definition $G'$ is a normal subgroup of $G$.
A: To prove that $G^{c}$ is a normal subgroup of $G$, it suffices to prove that if $(g,[c,d])\in{G}\times{G^{c}}$, then $g[c,d]g^{-1}\in{G^{c}}$. But, $$g[c,d]g^{-1} = [gcg^{-1}, gdg^{-1}].$$
Since $G^{c}$ is comprised by finite products of commutators in $G$, for any $x\in{G^{c}}$, it must be the case that $gxg^{-1}\in{G^{c}}$ for any $g\in{G}.$
A: if ${ghg^{-1} \in H}$ for all ${g \in G}$ and for all ${h \in H}$ , implies ${gHg^{-1} =H}$, then H is a normal subgroup.

*

*we only have to prove  that  ${g(xyx^{-1}y^{-1})g^{-1} \in H'}$ for all in G and x,y in G.
since ${gxyx^{-1}y^{-1}g^{-1}= ((gx)y(gx)^{-1}y^{-1}) \rightarrow (ygy^{-1}g^{-1}) \in H' }$, we are done

