- #1

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## 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##.