The normalizer of $H$ under the action of $G$ is
$$N_G(H) = \{g ∈ G | gHg^{−1} = H\}$$$S_3=\{(),(1,2),(1,3),(2,3),(1,2,3),(1,3,2)\}$
Let $G = S_3.$ Show that the normalizer of $\{1,(123),(132)\}$ is all of $S_3.$
First of all I guess $1$ and () are equal, it is just a notational difference(if it's not let me know)
Now I wanna ask how I should proceed to get all those elements in $S_3.$
I started as:
For $1\in H, g1g^{-1}=1 \in H ,\forall g\in G$
For $(123)\in H, 1(123)1^{-1}=(123) \in H $
For $(123)\in H, (12)(123)(12)^{-1}=(132) \in H$
For $(123)\in H, (13)(123)(13)^{-1}=(132) \in H$
For $(123)\in H, (23)(123)(23)^{-1}=(132) \in H$
For $(123)\in H, (123)(123)(123)^{-1}= (123) \in H$
For $(123)\in H, (132)(123)(132)^{-1}=(123)\in H$
I suppose I will get similar results for $(132)\in H \qquad$
as well. After showing that too am I done ? Are my computations valid? Or is there any better solution?
Also I wanna ask that is it true that $(12)^{-1}=(12) , (13)^{-1}=(13), (23)^{-1}=(23), (123)^{-1}=(123), (132)^{-1}=(132) $
I did this way on the above calculations
Best Answer
The way you did is one way but there are a lot more ways.
From the definition, if H is a subgroup of G, then the largest subgroup in which H is normal is the subgroup $N_G(H)$. If $H$ is itself normal in $G$ then $N_G(H) = G$. This way if you check (for finite group) that the order of $\frac{G}{H}$ is the smallest prime dividing $G$ then $H$ is normal in $G$. So $N_G(H) = G$.