Does it suffice to check the normal subgroup property for the generators?

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, $tat^{-1}\notin H$.

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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.

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 $.