# 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!

• With $h=t_1\ldots t_n$ and $g=s_1\ldots s_m$ observe $g^{-1}hg=g^{-1}t_1 g \ldots g^{-1}t_n g$. Since $H$ is closed under multiplication it is enough to check the case $h\in T$. But $g^{-1}hg=s_m^{-1}\ldots s_1^{-1} h s_1\ldots s_m$. Now $s_1^{-1} h s_1$ is in $H$ but we get stuck since we need it in $T$. – SK19 Apr 12 at 23:26
• Hint: Conjugation by an element $g\in G$ defines an automorphism $Inn_g$ of $G$. Now, prove that if $\phi: G\to G$ is an automorphism which sends generators of $H<G$ to elements of $H$ then $\phi(H)\subset H$. Lastly, analyze the relation between $Inn_g, Inn_f$ and $Inn_{gf}$. – Moishe Kohan Apr 12 at 23:38
• Ah right, this was the trick I was missing. Since $s_1^{-1}hs_1\in H$ we can write it again as $t'_1\ldots t'_{n_2}$ and start anew, effectively doing induction over $m$ (and within it, induction over $n_m$). – SK19 Apr 12 at 23:58
• @SK19: But remember that in order to generate a group you need to be able not only to multiply but also to invert, see Mike's answer. – Moishe Kohan Apr 13 at 0:03
• Yeah, I can See where I implicitly used that. I wondered if I had the right definition of generator on hand – SK19 Apr 13 at 6:23

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.

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 $$G= H$$.

• Just a remark: Bourbaki has a notion of “group with operators” and stable subgroup. For example, you can have a subset of G acting on the group by inner automorphisms. Modules are also an example of this structure. In this context, you can develop the notion – ACL Apr 20 at 18:19