Normaliser of a subset is a subgroup.

Question

I'm not sure where to start with this. I feel it should be fairly simple - maybe the notation is what is slipping me up.

I'm given $A$ is subset of $G$.

$N(A)$ is defined as $N(A)=\{x \in G : xA=Ax\}$ where $xA$, $Ax$ are the cosets as normal.

Prove $N(A) \le G$

Answer 1

We usually call $N(A)$ "the normalizer of $A$ in $G$." We can equivalently define it to be the set of all $x\in G$ such that $xAx^{-1}=A.$ Note that $A$ isn't assumed to be a subgroup of $G$, nor is it assumed that any of its cosets are subgroups of $G$. However, the normalizer of $A$ should be a subgroup of $G$, which is what you must prove.

What you should show is that:

• $1_G\in N(A),$ that is, $1_GA=A1_G.$
• If $x,y\in N(A),$ then $xy\in N(A)$, that is, if $xA=Ax$ and $yA=Ay$ then $xyA=Axy.$
• If $x\in N(A),$ then $x^{-1}\in N(A),$ that is, is $xA=Ax$ then $x^{-1}A=Ax^{-1}.$

It is worth noting that your title doesn't match the question at all, which makes me think you may be misunderstanding something, or leaving out some information. In particular, note that $\emptyset$ is not a subgroup of $G,$ but one can easily show (even more easily than using the above approach) that $N(\emptyset)=G.$