Showing normality by showing it holds for generators

[Mr Davis 97]

Homework Statement

Let $H = \langle S \rangle$ be a subgroup of $G = \langle T \rangle$. Prove that $H$ is normal in $G$ if and only if $tst^{-1} \in H$ for all $s \in S$ and $t \in T \cup T^{-1}$. Here $T^{-1}$ denotes the set $T^{-1}=\{t^{-1} \mid t\in T\}$.

Homework Equations

The Attempt at a Solution

The first direction is easy. Since $\forall g \in G$ we have $gHg^{-1} \subseteq H$ and since $S \subseteq H$ and $T \subseteq G$ then clearly $tst^{-1} \in H$ for all $s \in S$ and $t \in T \cup T^{-1}$.

The second direction is a bit more difficult. We first note that since $H = \langle S \rangle$ and $G = \langle T \rangle$, we know that if $h \in H$ then $h=s_1s_2\dots s_n$, where $s_i$ either represents an element in $S$ or the inverse of an element in $S$. Likewise, $g = t_1t_2\dots t_m$. Now, we want to show that $\forall n,m \in \mathbb{N}$, $ghg^{-1} = (t_1t_2\dots t_m)(s_1s_2\dots s_n)(t_m^{-1}\dots t_2^{-1}t_1^{-1}) \in H$ given that $tst^{-1} \in H$ for all $s \in S$ and $t \in T \cup T^{-1}$.

We first prove one lemma: $\forall s \in S$ and $\forall t \in T \cup T^{-1}$, $ts^{-1}t^{-1} \in H$. So, let $s \in S$ and let $t \in T \cup T^{-1}$ First, we know that $tst^{-1} \in H$. Now, $H$ is a subgroup and so is closed under inverses. So $(tst^{-1})^{-1} = t^{-1}s^{-1}t \in H$. But $t \in T \cup T^{-1}$, which means we also have the result $(t^{-1})^{-1}s^{-1}t^{-1} = ts^{-1}t^{-1} \in H$, which is what we wanted to show.

Now we want to proceed with the main result. Here is where I am sort of stuck. I can easily show that $\forall s \in S$ that $gsg^{-1}$ by induction on $n$ and that $\forall t \in T \cup T^{-1}$ that $tht^{-1}$ by induction on $m$, but I am not sure if this is sufficient to show that $\forall n,m \in \mathbb{N}$, $ghg^{-1} = (t_1t_2\dots t_m)(s_1s_2\dots s_n)(t_m^{-1}\dots t_2^{-1}t_1^{-1}) \in H$.

 

[fresh_42]

I think you cannot assume all powers $1$ on the left and all powers $-1$ on the right. You actually have to show
$$
t_1^{\varepsilon_1}\ldots t_m^{\varepsilon_m}\cdot s_1^{\eta_1}\ldots s_n^{\eta_n} \cdot t_1^{-\varepsilon_1}\ldots t_m^{-\varepsilon_m} \in H \; , \; \varepsilon_i,\eta_j = \pm 1
$$
which can be done in one single but long line by inserting $t_1^{\varepsilon_1}\ldots t_m^{\varepsilon_m}t_1^{-\varepsilon_1}\ldots t_m^{-\varepsilon_m}$ between all $s_j^{\eta_j}$ which is an essential step missing in your proof. It is actually the only step which is needed.

 

[Mr Davis 97]

I think you cannot assume all powers $1$ on the left and all powers $-1$ on the right. You actually have to show
$$
t_1^{\varepsilon_1}\ldots t_m^{\varepsilon_m}\cdot s_1^{\eta_1}\ldots s_n^{\eta_n} \cdot t_1^{-\varepsilon_1}\ldots t_m^{-\varepsilon_m} \in H \; , \; \varepsilon_i,\eta_j = \pm 1
$$
which can be done in one single but long line by inserting $t_1^{\varepsilon_1}\ldots t_m^{\varepsilon_m}t_1^{-\varepsilon_1}\ldots t_m^{-\varepsilon_m}$ between all $s_j^{\eta_j}$ which is an essential step missing in your proof. It is actually the only step which is needed.

So this is the only step that is needed. How do I prove that it holds for all positive integers $n$ and $m$? Do I use induction?

 

[fresh_42]

So this is the only step that is needed. How do I prove that it holds for all positive integers $n$ and $m$? Do I use induction?

Let's see. We have $tst^{-1}\in H$ for elements of $S^{\pm 1},T^{\pm 1}$ resp. We need $ghg^{-1} \in H$, but $g=t_1^{\varepsilon_1}\ldots t_m^{\varepsilon_n}$ and $h=s_1^{\eta_1}\ldots s_m^{\eta_m}$. I have been way too optimistic in the previous post, since I hadn't considered, that $tst^{-1} {\notin}_{i.g.} S^{\pm 1}$, which indeed complicates the case. The problem is that $tst^{-1}={s'\,}_{1}^{\tau_{1}}\ldots {s'\,}_{n_1}^{\tau_{n_1}}$ and every insertion of another $t_it_i^{-1}$ might blow up the $s-$word and we possibly will never come to an end. To confine this situation, an induction might help. But I'm not sure and not whether parallel or serial inductions will be necessary. I hope first $n$ and then $m$ will do, for otherwise it will become confusing.