4
$\begingroup$

Let $S_4$ denote the group of permutations of $\{1,2,3,4\}$ and let $H$ be a sub group of $S_4$ of order $6$ . Show that $\exists~ i \in \{1,2,3,4\}$ which is fixed by each element of $H$.

Attempt: As per the given question, $H$ is a sub group of $S_4$ of order $6$ .

We have to prove that $\exists~ i \in \{1,2,3,4\} ~ s.t. ~ \alpha(i) = i ~\forall ~\alpha \in H$ . So, If i prove that $ H \subseteq Stab_G (i)$ , then I can prove the same.

Now, by orbit stabilizer theorem, we have $|G| = |Stab_G (i)|~|Orb_G(i)|$ .

{ So, if $G$ is a group of permutation of a set $S$. For $\forall ~s \in S$, $orb_G(s)= \{\phi(s)~ | ~~\phi \in G .$ The set $orb_G(s)$ is a subset of $S$ called the orbit of $s$ under $G$. }

We also know that $Stab_G(i)$ is a subgroup of $G$.

IF i show that $\exists ~orb_G(i)$ such that $| orb_G(i) |=4$ under $G$, then we are done because $Stab_G (i)$ is already a subgroup of $G$.

So, my problem boils down to finding an $i$ such that $| orb_G(i) |=4$

EDIT : I also know that Since $S_4$ has no elements of order $6$, by Lagrange's theorem the elements of H must have orders $2$ or $3 $ which means they can be which are $2$ cycled or $3$ cycled

How do I find such an $i$. Do i now write all the elements of $S_4$? Writing down everything seems tedious and impractical though ? Help will be really appreciated.

Thank you.

$\endgroup$
5
$\begingroup$

Since $S_4$ has no elements of order $6$, the elements of $H$ must have orders $2$ and $3$ (and of course $1$ for the neutral eleement). Especially, $H$ has an element of order $3$, which must be a 3-cycle $(a\,b\,c)$. (It thenalso contains $(a\ c\,b)$). Let $d$ be the one element that does not belong to this 3-cycle. Also, $H$ has an element of order $2$, which can be either of the form $(x\,y)$ or $(x\,y)(u\,v)$. In the first case, if $d\notin \{x,y\}$, we are done. Otherwise note that e.g. $(a\,b\,c)(a\,d)=(a\,d\,b\,c)$ is of order $4$ and cannot be $\in H$. For the other form, note that e.g. $(a\,b\,c)\cdot (a\,b)(c\,d)=(a\,c\,d)$ and $(a\,c\,b)\cdot (a\,b)(c\,d)=(b\,c\,d)$, whence we find at least $6$ elements of order $3$ in $H$, which is impossible.

$\endgroup$
  • 1
    $\begingroup$ @VHP: In this case, it's quite easy to show that if a subgroup's non-identity elements all have order 2, then the subgroup has order 2 or 4, not 6. Similarly, if all non-identity elements have order 3, then the subgroup has order 3. $\endgroup$ – Alastair Litterick Apr 10 '14 at 23:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.