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

1. Jul 27, 2015

### 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. The problem statement, all variables and given/known data

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.

3. 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.

2. Jul 27, 2015

### Zondrina

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 $\left< S \right> \unlhd G$.

3. Jul 30, 2015

### NihilTico

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.

4. Jul 31, 2015

### 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 = Ug ∈ Gg−1Tg. Show that ⟨S⟩ is the normal subgroup generated by T.

5. Jul 31, 2015

### NihilTico

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].

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?