Nontrivial normal subgroup that doesn't contain commutator subgroup

Can there be non-trivial normal subgroups that do not contain the commutator subgroup $$C$$?

One can show that any subgroup $$H$$ that contains $$C$$ is a normal subgroup with a few algebraic manipulations. However, I am curious about whether it is also true that every normal subgroup must contain the commutator subgroup. I started with the definition of a normal subgroup, that the left and right cosets are equal, and it follows that it is invariant under conjugation. As such, $$gh_1g^{-1} = h_2$$ for $$h_1, h_2 \in H$$, and therefore the commutator $$gh_1gh^{-1}h^{-1} \in H$$, but that isn't the set of all commutators because our conjugation process doesn't account for elements not in $$H$$. Is there a way to show that all commutators are part of any normal subgroup?

• Mar 21 '19 at 17:06
• The commutator subgroup of $S_n$ is $A_n$. Does $S_4$ contain any normal subgroups other than $\{e\}$, $A_4$, and $S_4$? Or, if $G$ is not abelian, and $A$ is abelian, what is commutator subgroup of $G\times A$? And is $\{e\}\times A$ normal in $G\times A$? And for one with partial intersection, take group $G_1$ and $G_2$ with $1\neq [G_1,G_1] \neq G_1$, $1\neq [G_2,G_2]\neq G_2$, and look at $G_1\times G_2$. Mar 21 '19 at 17:18

So to find a counterexample, all you have to do is to define a homomorphism from a group $$G$$ onto a nonabelian group $$H$$.
So, take your favorite nonabelian group, I'll take $$S_3$$, the symmetric group on three symbols. Define $$G = S_3 \times S_3$$, define $$H = S_3$$, and define the surjective homomorphism $$f : G \to H$$ by $$f(a,b)=b$$. Its kernel is a nontrivial normal subgroup of $$G$$ that does not contain the commutator subgroup of $$G$$.