If the intersection of $S$ and $N$ is normal in $N$, then $S$ is normal in $SN$.

Question

Let $G$ be a group, let $N$ be a normal subgroup of $G$ and let $S$ be a subgroup of $G$. If $S \cap N$ is normal in $N$ and $N / S\cap N$ is free abelian, is $S$ normal in $SN$?

I guess this is not true, but does $SN / S$ have a group structure? If I consider the set of left cosets of $SN$ in $S$, there is a still a bijection $$\{ g S \cap N \mid g\in N \} \mapsto \{g S \mid g\in SN\},$$ so can we define an operation in $SN/ S$ based on that bijection?

Answer 1

In general, no, you don't know if $S$ is normal in $SN$. Here's an example: take $N=\mathbb{Z}\times\mathbb{Z}$, $S=C_2$, the cyclic group of order $2$ generated by $x$, and let $S$ act on $N$ by exchanging coordinates. Then let $G=N\rtimes S$.

Then $S\cap N$ is trivial, which is normal in $N$; $N/(S\cap N)=N$ which is free abelian. $SN=G$. But $S$ is not normal in $G$, since for instance $$ \bigl((1,0),e\bigr)^{-1}\bigl((0,0),x\bigr)\bigl( (1,0),e\bigr) = \bigl( (-1,0),e\bigr)\bigl( (0,1),x\bigr) = \bigl( (-1,1),x\bigr)\notin S.$$

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You are giving a bijection between the cosets of $S\cap N$ in $N$ and the cosets of $S$ in $SN$ by identifying the coset $n(S\cap N)$, $n\in N$, with the coset $nS$. This works, in that $n(S\cap N)=m(S\cap N)$ if and only if $m^{-1}n\in S\cap N\subseteq S$, hence $nS = mS$. Conversely, if $nS=mS$ then $m^{-1}n\in S$ and $m^{-1}n\in N$, so $n(S\cap N)=m(S\cap N)$. Thus the map is injective. And finally, any coset of $S$ in $SN$ is of the form $(sn)S$ for some $s\in S$, $n\in N$, which we can rewrite as $(sn)S = (sns^{-1})sS = (sns^{-1})S$, and $sns^{-1}\in N$, so every coset is of this form.

So you can certainly use transport of structure to give the cosets of $S$ in $SN$ the same group structure as $N/(S\cap N)$. The problem is that this group structure is not particularly natural.

Why? Because we don't get that $(xS)(yS)$ is necessarily equal to $xyS$ when $x,y \in SN$. Write $x=sn$, $y=s'n'$. Then as above, we have that $x$ corresponds to the coset $sns^{-1}(S\cap N)$, and $y$ corresponds to the coset $s'n'(s')^{-1}(S\cap N)$, so the "product" we are defining using transport of structure is that $$(xS)\odot(yS) = \Bigl( (sn)S\Bigr)\odot\Bigl((s'n')S\Bigr) = (sns^{-1})(s'n'(s')^{-1})S.$$ By contrast, $xyS = (sn)(s'n') = (ss')\bigl((s')^{-1}ns'\bigr)n'$, so this corresponds to the coset $(s')^{-1}ns'n'S$. And in general, you do not have $$(sns^{-1})(s'n'(s')^{-1})S = (s')^{-1}ns'n'S.$$ So the group structure you are giving to the cosets of $S$ in $SN$ is not the "natural" one, which probably makes it not particularly useful. You might as well give it a group structure by bijecting with a suitable free abelian group.

Answer 2

Let $$ G= \left(% \begin{array}{cc} \pm1 & \mathbb{Z} \\ 0 & \pm1 \\ \end{array}% \right),\ N= \left(% \begin{array}{cc} 1 & \mathbb{Z} \\ 0 & 1 \\ \end{array}% \right),\ S= \left(% \begin{array}{cc} \pm1 & 0 \\ 0 & \pm1 \\ \end{array}% \right). $$ Then $N$ is a normal subgroup of $G$, $NS=G$, $S\cap N=\{e\}$, and $N/S\cap N\cong\mathbb{Z}$. However, $S$ is not a normal subgroup in $G$. Further, if $g\in N$, then $gS\cap N=\{g\}$ and the operation on $SN/S$ can be defined, but whether it will be useful is not clear.