Is $G$ normal subgroup of $S_6$?

Question

Let $G \subset S_6$ will be subgroup of permutation of a set $I_6=\{1,2,3,4,5,6\}$ such that $$\forall x \in I_6 :2|x \Longleftrightarrow 2|\sigma(x).$$

Is $G$ is normal subgroup of $S_6$? I am a little confused, because I am not sure how it (construction of group $G$) 'works'.

Answer 1

The subgroup $G$ (closure) is made of the permutations on $I_6:=E\sqcup O$ ("even and odd") which send "even-to-even" and (then) "odd-to-odd". Take an $\alpha\in S_6$ which sends some even to odd, but not all the even to the odd, say it "mixes up even and odd"; then $\alpha(E)$ is neither $E$ nor $O$. Accordingly, we can find a permutation $\sigma\in G$ which sends $\alpha(E)$ to a set distinct from $\alpha(E)$: for example, legitimate such $\alpha$'s and $\sigma$'s are all those such that:

$$\alpha(2)=\color{red}{1}, \space\space\space\alpha(4)=\color{blue}{4}, \space\space\space\alpha(6)=\color{green}{2}\tag 1$$

and

$$\sigma(\color{red}{1})\space\stackrel{odd-to-odd}{=}\space3, \space\space\space\space\sigma(\color{blue}{4})\space\stackrel{even-to-even}{=}\space2, \space\space\space\space\sigma(\color{green}{2})\space\stackrel{even-to-even}{=}\space4$$

So, by choosing "mixing up" $\alpha$'s ("exists..."), we have been able to build up $\sigma$'s ("exists...") such that:

$$\sigma(\alpha(E))=\{3,2,4\} \ne \{1,4,2\}=\alpha(E)=\alpha(\tau(E)), \space\forall \tau\in G \tag 2$$

We can read $(2)$ this way:

$$\exists\alpha\in S_6, \exists \sigma\in G\mid (\sigma\alpha)(E)\ne (\alpha\tau)(E), \space\forall \tau \in G \tag 3$$

whence:

$$\exists\alpha\in S_6, \exists \sigma\in G\mid \alpha^{-1}\sigma\alpha\ne \tau, \space\forall \tau \in G \tag 4$$

or equivalently:

$$\exists\alpha\in S_6, \exists \sigma\in G\mid \alpha^{-1}\sigma\alpha\notin G \tag 5$$

meaning $G\ntrianglelefteq S_6$.

Answer 2

Take an element in $G$, for instance $$ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 4 & 3 & 2 & 5 & 6 \end{pmatrix} $$ As a product of disjoint cycles it is the transposition $(24)$. Let's conjugate by $(12)$: $$ (12)(24)(12)=(14) $$ which does not belong to $G$, because it maps $4$ to $1$.

Counterexample found, the subgroup is not normal. By the way, I knew it isn't normal even before looking for the counterexample, because the only nontrivial normal subgroup of $S_6$ is $A_6$ (which is known to be a simple group) and certainly $G\ne A_6$, because $(123)\in A_6$, but $(123)\notin G$.