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.


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.

  • 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

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