Show that a normal subgroup <S> is equal to <T>

[QIsReluctant]

Note: I only need help on the underlined portion of the problem, but I'm including all parts since they may provide relevant information. Thanks in advance.

1. Homework Statement
Let S be a subset of a group G such that g⁻¹Sg ⊂ S for any g∈G. Show that the subgroup ⟨S⟩ generated by S is normal.
Let T be any subset of G and let S = g⁻¹Tg. Show that ⟨S⟩ is the normal subgroup generated by T.

The Attempt at a Solution

I apply the first part of the problem to see that <S> is normal, and that is as far as I am getting. I know that <S> ⊇ <T> by the definitions, but since we have such little information about T I can't get much further. If the normal/"ordinary" subgroups containing S and T were the same then the conclusion would be obvious but the definition of G seems to preclude this.

 

[STEMucator]

Let S be a subset of a group G such that g−1Sg ⊂ S for any g∈G. Show that the subgroup ⟨S⟩ generated by S is normal.

So $S \subseteq G$, and $g^{-1} S g \subset S, \forall g \in G$.

Did you mean $S \subseteq G$, and $g^{-1} S g = S, \forall g \in G$?

This would tell you $S$ is a normal subgroup of $G$ because the similarity transformation holds $\forall g \in G$.

Let $\left< S \right>$ be the subgroup generated by $S$. Then $\left< S \right> \subseteq S \subseteq G, \forall g \in G$ and $\left< S \right>$ is also a subgroup of $G$. The similarity transformation of $\left< S \right>$ by some $g \in G$ that is not in $\left< S \right>$ will always give a subgroup of $G$.

So all that would be left to show is $g^{-1} \left< S \right> g = \left< S \right>, \forall g \in G$. You know every element in $\left< S \right>$ is contained in $S$ and $S$ is normal. So $\left< S \right> \unlhd G$.

 

[NihilTico]

Let T be any subset of G and let S = g⁻¹Tg. Show that ⟨S⟩ is the normal subgroup generated by T.

If by this, you mean that $\langle gTg^{-1}\rangle=\langle{T}\rangle$ for any subset of a group, then I dispute your claim.

Just consider a group $G$ where $G'$ is a subgroup of $G$ that is not normal (i.e., $gG'g^{-1}\ne G$). Indeed, if the set $\{g_1,\ldots,g_n\}$ generate $G'$, then their conjugates $\{gg_1 g^{-1},\ldots,gg_{n}g^{-1}\}$ obviously generate $gG'g^{-1}$. This is a counter example to your claim that $\langle gTg^{-1}\rangle=\langle T\rangle$ for any subset of $G$. So I'm not entirely sure what you're getting at with the underlined portion of your question.

 

[QIsReluctant]

Correction!
Let S be a subset of a group G such that g⁻¹Sg ⊂ S for any g∈G. Show that the subgroup ⟨S⟩ generated by S is normal.
Let T be any subset of G and let S = U_{g ∈ G}g⁻¹Tg. Show that ⟨S⟩ is the normal subgroup generated by T.

 

[NihilTico]

Correction!
Let S be a subset of a group G such that g⁻¹Sg ⊂ S for any g∈G. Show that the subgroup ⟨S⟩ generated by S is normal.
Let T be any subset of G and let S = U_{g ∈ G}g⁻¹Tg. Show that ⟨S⟩ is the normal subgroup generated by T.

I think I misinterpreted your question the first time around. Do you mean by 'the normal subgroup generated by $T$' the normal closure of $T$? The normal closure of $T$ is the minimal normal subgroup containing $T$.

If so, then let's claim that $S'=\langle\bigcup_{g\in{G}}gTg^{-1}\rangle$ is the minimal such group. Then, notice this means that $S'=\bigcap\{N\colon N\unlhd G\text{ and }T\subseteq{N}\}$, which is to say, $S'$ is the intersection of all normal subgroups containing $T$. That the intersection of all normal subgroups is the minimal such normal subgroup containing $T$ is clear, for if there were another, then it was in our intersection.

[As an exercise, show that the intersection of an arbitrary collection of normal subgroups of a group is normal]. Here's how I would go about this problem:

First, notice that our claim that $S'$ is the minimal such normal subgroup containing $T$ is true $\iff$ $S'$ is contained in every normal subgroup that contains $T$, after all, $S'$ is a normal subgroup containing $T$.

[Pf: $(\implies)$ If $S'$ is the minimal such normal subgroup, then $S'$ is in the intersection $S''=\bigcap\{N\colon N\unlhd G\text{ and }T\subseteq{N}\}$ as $T\subseteq S'$ and $S'\unlhd G$; hence, $S''\subseteq S'$ (this is just a property of intersections). By the exercise above, $S''\unlhd G$ and since $S'$ is in this intersection, $S''\unlhd S'$, but this contradicts the minimality assumption on $S'$, so $S''=S'$, and so $S'$ is contained in every member of the intersection. $(\impliedby)$ This is straight forward because $T\subseteq S'$, $S'\unlhd G$ and $S'$ is contained in every member of the intersection of $S''$.]

Therefore, it suffices to prove that $S'$ is contained in every normal subgroup that contains $T$. Now it is straight forward. Do you see why?