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Let $G$ be a group and $S,T \subset G$ are subsets of $G$ such that $\langle T\rangle =G$ and $\langle S \rangle=N$. Does the following hold:

$N$ is normal in $G$ iff for all $t\in T$ we have $tSt^{-1} \subseteq N$

Here $\langle X \rangle$ denotes the subgroup generated by $X$ (the smallest subgroup of $G$ containing $X$)

I only have the answer for $G$ is finite group.

Edit: Sorry, now the question is right.

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You have the answer when $G$ is finite... what is it? – Mariano Suárez-Alvarez Apr 11 '12 at 2:11
I think that you must have reversed $S$ and $T$: surely you want the condition to be that $sTs^{-1}\subseteq N$ for all $s\in S$. – Brian M. Scott Apr 11 '12 at 2:19
Perhaps I’m overlooking something obvious, but from $S^T\subseteq N$ don’t you have almost immediately that $N^T\subseteq N$, and thence that $N^G=N$? – Brian M. Scott Apr 11 '12 at 2:25
@BrianM.Scott I can't see that coming, could you elaborate more? – Ajat Adriansyah Apr 11 '12 at 2:31
@MarianoSuárez-Alvarez It is true for finite group, since then the element of $G$ is all finite product of element of $T$ and the element of $N$ is all finite product of element of $S$. For infinite group, the element of $G$ is $t_1 t_2 \cdots t_k$ where either $t_i \in T$ or $t_i^{-1} \in T$, I can't conclude because the inverse might not be in $T$. – Ajat Adriansyah Apr 11 '12 at 2:32
up vote 3 down vote accepted

The answer to your question is "no" in general. It is not enough in general.

Here's an example: we will work inside the group of invertible $2\times 2$ matrices with rational coefficients. Let $N$ be given by $$N = \left\{\left.\left(\begin{array}{cc} 1 & m\\ 0 & 1\end{array}\right)\in G\ \right|\ m\in\mathbb{Z}\right\},$$ and let $$x = \left(\begin{array}{cc} 2 & 0\\ 0 & 1\end{array}\right).$$ Let $G = \langle N,x\rangle$. Set $T=N\cup\{x\}$, and set $S=N$. Then trivially $N=\langle S\rangle$, and $G = \langle T\rangle$. Also, for every $n\in N$ we have $nNn^{-1}\subseteq n$ (you can verify that this is the case); and $xNx^{-1}\subseteq N$, since for every $m\in \mathbb{Z}$, we have: $$\begin{align*} \left(\begin{array}{cc} 2 & 0\\ 0 & 1\end{array}\right)\left(\begin{array}{cc}1 & m\\0 & 1\end{array}\right) \left(\begin{array}{cc}\textstyle\frac{1}{2}&0\\0&1\end{array}\right) &= \left(\begin{array}{cc} 2 & 2m\\0 & 1\end{array}\right)\left(\begin{array}{cc} \textstyle\frac{1}{2}&0\\0 & 1\end{array}\right)\\ &= \left(\begin{array}{cc} 1 & 2m\\ 0 & 1\end{array}\right)\in N. \end{align*}$$

However, $N$ is not normal, since $$\begin{align*} x^{-1}\left(\begin{array}{cc}1&1\\0&1\end{array}\right)x &= \left(\begin{array}{cc} \textstyle\frac{1}{2} & 0\\0 &1\end{array}\right)\left(\begin{array}{cc}1 & 1\\ 0 & 1\end{array}\right)\left(\begin{array}{cc}2 & 0\\0&1\end{array}\right)\\ &= \left(\begin{array}{cc} \textstyle\frac{1}{2}&\textstyle\frac{1}{2}\\0&1\end{array}\right)\left(\begin{array}{cc}2&0\\0&1 \end{array}\right)\\ &= \left(\begin{array}{cc} 1 & \textstyle\frac{1}{2}\\0&1\end{array}\right)\notin N. \end{align*}$$ So we have $tSt^{-1}\subseteq N$ for all $t\in T$, but $N$ is not normal.

However, if $T$ is a generating set and $tNt^{-1}=N$ for all $t\in T$, then the conclusion holds. In particular, this gives your result in the case where $N$ is finite, since the hypothesis imply that $tNt^{-1}\subseteq N$ for all $t\in T$, hence (by finiteness of $N$), that $tNt^{-1} = N$, which in turn gives you $t^{-1}Nt=N$ as well.

The difficulty arises because in an infinite group, given a subgroup $N$, the subset $$\{ g\in G\mid gNg^{-1}\subseteq N\}$$ need not be closed under inverses, as noted in this previous question from whence the example came.

share|cite|improve this answer
ahaa, nice :D ... I was also messing around with $$\left(\begin{array}{cc} 1 & m\\ 0 & 1\end{array}\right)$$ , but I didn't put the inverse of this matrix in my generating set. thanks anyway, neat example indeed :D – Ajat Adriansyah Apr 11 '12 at 9:25

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