For a Group $G$ with$S\subseteq G$, when does $ \left\langle S \right\rangle $ equal the normal closure of $S$?

Question

To obtain the group with the presentation that has all elements of a subset $S$ equal to the identity, you must take the quotient of the free group (on the appropriate number of generators) that sets the $normal$ $closure$ $of$ $S$ to the identity. I notice that in the case of the abelianization of G, this is equivalent to setting the group generated by the commutators of G to the identity. This is the case where S is the set of all commutators of G. My question is, when is this true in general? When can we calculate the normal closure of a set in a group just by looking at the subgroup generated by the set?

Answer 1

In general: It is not possible to determine whether a subgroup $H$ of a group $G$ is normal just by looking at $H$ (or at any generating set, for that matter).

Let $H=\langle S\rangle$. First, let's show the following are equivalent:

1. For every group $G$ containing $H$ and for every $\sigma\in\operatorname{Aut}(G)$, we have $\sigma(H)=H$.
2. For any group $G$ containing $H$, $H$ is normal in $G$.

Suppose 1., and let $G$ be a group containing $H$. Let $g\in G$. Then conjugation by $g$, i.e., the map $x\mapsto gxg^{-1}$, is an automorphism of $G$ and thus preserves $H$, i.e., $gHg^{-1}=H$. Therefore $H$ is normal.

Now suppose 2., and let $G$ be a group containing $H$. Let $\sigma\in\operatorname{Aut}(G)$. Consider the new group $K=G\rtimes\operatorname{Aut}(G)$, which contains $G$, and thus $H$, and the action of $\sigma$ on $G$ becomes conjugation by $\sigma$ in $K$. In particular, $H$ is normal in $K$, so $\sigma(H)=\sigma H\sigma^{-1}=H$.

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Now let $H$ be any group, consider $G=H\times H$, and look at $H$ as $H\times 1$ in $G$, and the automorphism $\sigma:(x,y)\mapsto (y,x)$ of $G$. Then $\sigma$ does not preserve $H$.

Thus, 1. above is false, and therefore 2. is also false.