Cfrgtkky

Let $ Tsubseteq S$ denote a subset of $S$ and $G$ denote the group acting on $S$. Let $$G_T=bigcap_{sin T} G_s quadtext{and}quad G_{{T}}={ gin G: T = T^g } $$
represent the pointwise stabilizer and setwise stabilizer, respectivily (note that I'm using the exponential notation for a group action, and that $G_s$ represents the stabilizer of $s$).

I tried to prove three different things:

1. 
   $G_T$ is a subgroup of $G$
   


2. 
   $G_{{T}}$ is a subgroup of $G$
   


3. $G_T trianglelefteq G_{{T}}$

For 1, I used the fact that the intersection of subgroups is also a subgroup. For 2, I actually thought of the one-step subgroup test, but I don't know how to do this (maybe first proving that for each element in $G_{{T}}$, the inverse element is also included in $G_{{T}}$).
For 3, I tried to prove that $forall s in G_T, g in G_{{T}}: g^{-1} s g in G_T$, but that didn't work out either.

Can you give me the proof of (2.) and (3.)? I've searched a lot on the Internet, but didn't found anything (and as you probably could have guessed by this moment, I'm a beginner). Thank you for considering my request!

For the second part, there is an easy way to see that the result is true : $T^{gh} = (T^g)^h = T^h = T$, if $g,h$ are in $G_{{T}}$. If you do not immediately why this sequence of equalities holds, you should try to convince yourself that it is true (and not just try to find another solution). A similar argument shows that $T^{g^{-1}} = T$ if $g$ is in $G_{{T}}$.

For the third part, do not forget to first prove that $G_T$ is indeed a subgroup of $G_{{T}}$ before proving that it is a normal subgroup ! This part is not too hard but you need to do it.
You were on the right track though : to prove $g^{-1}sg$ is in $G_T$, you just need to prove that for every $t in T$, $t^{g^{-1}sg} = t$. Just write $t^{g^{-1}sg} = ((t^{g^{-1}})^s)^g$ and you should be able to prove it.

Also, you should try developping some intuition for the last result, which is actually really natural : $g^{-1}$ "shuffles" your set $T$, then $s$ acts like the identity on it, then $g$ cancels the first "shuffling". So in the end, $g^{-1}sg$ acts like the identity on $T$, which is the definition of an element of $G_T$.

This notion of shuffle can be seen more easily with the first "simplest" group action there is : the symmetric group $mathfrak{G}_n$ acting on ${1,dots,n}$. Take for example the case $n = 4$, $T={1,3}$, and try to write what $G_T$ and $G_{{T}}$ actually are, and compute an example of $g^{-1}sg$ in this setting. This should (in my opinion) be really enlightening.