Does it suffice to check the normal subgroup property for the generators? Let $G$ be a group generated by a subset $S$ and $H$ be a subgroup of $G$ generated by a subset $T$.
To check whether $H$ is a normal subgroup of $G$ or not, we must check the following statement:
$$
\forall g \in G \: \forall h \in H: \: g^{-1} h g \in H.
$$

Question: Does it suffice to check $$
\forall s \in S \: \forall t \in T: \: s^{-1} t s \in H?
$$

I assume that this is true, but the proof of that seems to be really technical. Could you please help me by answering and explaining my question? 
Any help is really appreciated!
 A: Here's a proof of Mike's modified  statement without induction, using only the definition of generating subset (I find it cleaner this way)
Fix $ t \in T $. Then the set of $ g $ s such that $ g t g^{-1} \in H $ is closed under multiplication, and contains $ S \cup  S^{- 1} $. So it contains the submonoid generated by this set, which, by similar methods, is easily seen to be $ G $. 
Therefore for all $ g \in G $ , $ gtg^{-1} \in H $. 
This is true for all $ t \in T $ , and the set of $ x $ for which it is true is clearly closed under multiplication and inverses therefore it must contain $ H $. Thus for all $ g \in G, gHg^{-1} \subset H $, which is all we wanted. 
Appendix : if $ S $ generates $G $ then $ S \cup S^{-1} $ generates $ G $ as a monoid. Indeed let $ H $ be the monoid generated by this set. The set of $ x \in H$ such that $ x^{-1} \in H $ contains $ S $ by construction and is closed under multiplication  (because $ H $ is) and under inverses . Therefore it is $ G $. But it is included in $ H $, therefore $ G= H $.
A: No, it does not always suffice. Consider the Lamplighter group. This has two generators, $a$ and $t$, representing transformations of functions $f:\mathbb Z\to \{0,1\}$. 


*

*$a$ changes the value of $f(0)$, and leaves all others the same.

*$t$ shifts the sequence by one, replacing $n\mapsto f(n)$ with $n\mapsto f(n+1)$.
Let $H$ be the subgroup generated by $a,t^{-1}at^{},t^{-2}at^{2},\dots$  You can verify that $t^{-1}Ht\subseteq H$, and $a^{-1}Ha=H$. Since $a,t$ generates the group, your condition would imply $H$ was normal. However, $tat^{-1}\notin H$. 

However, this modified statement is true.

If $S$ generates $G$ and $T$ generates $H$, and $\forall s\in S,t\in T$ we have
  \begin{align}s^{-1}ts\in H\quad \text{and} \quad sts^{-1}\in H,\end{align} then $H$ is normal in $G$.  

Proof The condition further implies $s^{-1}t^{-1}s=(s^{-1}ts)^{-1}\in H$ as well. 
Next, for all $s\in S$, $h\in H$, we have $s^{-1}hs\in H$ and $shs^{-1}\in H$. To see this, write $h=t_1t_2\dots t_n$ with each $t_i\in T$ or $t_{i}^{-1}\in T$. Then
$$
s^{-1}hs=(s^{-1}t_1s)(s^{-1}t_2s)\cdots (s^{-1}t_ns)\in H
$$
since all factors are in $H$. The same goes for $shs^{-1}$.
Now, given $g\in G$, $h\in H$, we can write $g=s_1s_2,\dots,s_n$, where either $s_i\in S$ or $s_i^{-1}\in S$. Now, define a sequence $h_0,h_1,\dots, h_n$ by


*

*$h_0 = h$.

*$h_{i+1} = s_{i+1}^{-1}h_{i} s_{i+1}$ for $i=0,1,2,\dots,n-1$.
We can prove by induction, and using the facts $s^{-1}hs\in H$ and $shs^{-1}\in H$ for all $h\in H$, that $h_{i}\in H$ for each $i$. But $h_n=s_n^{-1}\dots s_2^{-1}s_1^{-1}hs_1s_2\dots s_{n}=g^{-1}hg$,
so we are done.
