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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?

Arturo Magidin
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Tarang Saluja
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  • Related (to your question). – Shaun Mar 21 '19 at 17:06
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    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 $G_1\times G_2$. – Arturo Magidin Mar 21 '19 at 17:18

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If every nontrivial normal subgroup contained the commutator subgroup then every quotient group modulo a nontrivial normal subgroup would be abelian.

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 $G$.

Lee Mosher
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