-2

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'.

pawelK
  • 457
  • 2
    I see that you've been here for over a year, so surely you know that you should avoid "no clue" questions like this. (Also, it is unclear what you mean by "it". Do you mean proving normality, or do you mean how the subgroup is defined?) – user1729 Jul 07 '20 at 16:39
  • @PeterForeman This is permutation value for $x$. \begin{pmatrix} 1 & 2 & \cdots \ \sigma(1) & \sigma(2) & \cdots \end{pmatrix} – pawelK Jul 07 '20 at 16:40
  • If $s\in G$ then $s$ maps $A={1,3,5}$ to $A$ and $s$ maps $B={2,4,6}$ to $B$ so $s=s_As_B=s_Bs_A$ where $s_A$ is a permutation of $A$ and $s_B$ is a permutation of $B.$ – DanielWainfleet Jul 07 '20 at 17:19
  • https://math.stackexchange.com/questions/482889/normal-subgroups-of-s-n-for-n-geq-5 . From my prior comment, $G$ has $3!^2=36$ members but $A_6$ has $6!/2=360$ members. – DanielWainfleet Jul 07 '20 at 17:25
  • 1
    I notice that you addressed the second part of my comment, about what "it" is. However, your question is still a "no clue" question. You need to improve it. – user1729 Jul 07 '20 at 18:30
  • 1
    Try conjugating by the first element you think of? Say, $(1,2)$? – David A. Craven Jul 07 '20 at 19:00
  • @runway44 I know it's not in $G$. Conjugating by something that is already in $G$ wouldn't be very useful. – David A. Craven Jul 07 '20 at 22:51
  • @runway44 Yes exactly, you want to conjugate something in to get something not in to prove it's not normal. But for that, you need to choose a conjugating element that is not in $G$. – Derek Holt Jul 08 '20 at 07:51
  • Oh. Somehow I missed the word "by" in David's comments. – anon Jul 08 '20 at 08:12
  • Aha. Yes, conjugating $(1,2)$ wouldn't do much good, true! – David A. Craven Jul 08 '20 at 08:35

2 Answers2

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 $G\ntrianglelefteq S_6$.

1

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 $(123)\notin G$.

egreg
  • 238,574