Prove that $H$ is a normal subgroup in the finite group $G=\langle S\rangle $ iff $sHs^{-1}$ $\subseteq H$ for all $s\in S$. 
Let H be a subgroup in a group $G$. If $G$ is generated by a set $S$, so $G=\langle S\rangle $, prove that $H$ is a normal subgroup in $G$ iff $sHs^{-1}$ $\subseteq H$ for all $s\in S$. 

Note with the question: Results of this sort are useful because it is easier to check an algebraic property for a small set of generators than to prove it holds for all elements of G.
Hi, the forward direction is quite straightforward but I will be really grateful if somebody can help me with the reverse direction. Thanks!
 A: It seems this is wrong. The problem is that $S$ does not need to be closed under inverses.
Let $N = \oplus_{i \in \mathbb{Z}} \mathbb{Z}_2$ and let $K = \langle g \rangle \cong \mathbb{Z}$ act on $N$ by $g \mapsto$ shifting right by 1 unit. Form the semidirect product $G = N \rtimes K$. Let $M$ be the subgroup of $N$ consisting of sequences with 0 entries in negative coordinates. Let $S = N \cup \{g\}$. Then $S$ generates $G$ and fixes $M$.
But $M$ is not normal in $G$.
A: Just collecting some ideas from the long thread:
The statement IS true if we have $sHs^{-1} = H$ for all $s \in S$.
Proof: Consider $N_G(H) = \{g \in G \mid gHg^{-1} = H\}$, this is called the normalizer of $H$ in $G$. You can prove (for arbitrary $G$ and $H$) that this is a subgroup of $G$. In our case, our assumption is that $S \subseteq N_G(H)$, and so $\langle S \rangle \subseteq N_G(H)$, and so $N_G(H) = G$. This implies $H$ is normal in $G$.
Now, suppose we only know that $sHs^{-1} \subseteq H$. We can finish the proof if:
$\bullet$ $G$ is finite: In this case, $H$ is finite, and $sHs^{-1}$ has the same number of elements as $H$, so we must have $sHs^{-1} = H$.
$\bullet$ $S$ contains inverses: In this case, for all $s \in S$ we have $sHs^{-1} \subseteq H$ and $(s^{-1})H(s^{-1})^{-1} = s^{-1}Hs \subseteq H$. Together, these imply that $sHs^{-1} = H$.
As others have shown, the statement is NOT true as you have written it.
A: Write down an arbitrary element in $G$ in terms of the generating set.  See what happens when you apply the element to $H$ by conjugation.
