1
$\begingroup$

I have to find the subgroups of $S_4$ of order 6:

<(12),(123)>={1,(12),(123),(132),(23),(13)}

but how much are?

maybe 4 : <(12),(124)>={1,(12),(124),(142),(24),(14)}

      <(13),(134)>={1,(13),(134),(143),(34),(14)}

      <(23),(234)>={1,(23),(234),(243),(34),(24)}
$\endgroup$
1
  • $\begingroup$ Do you know all the different types of group of order 6? Then you can work out how each type can sit in $S_4$, and then you can work out how many of each type there are in $S_4$. $\endgroup$ – Gerry Myerson Jan 28 '16 at 11:32
1
$\begingroup$

There are two possible groups of order $6$: $C_6$ and $S_3$.

Since $S_4$ has no element of order $6$, the only possibility is a subgroup isomorphic to $S_3$, and these are the conjugates of $S_3$ in $S_4$.

$\endgroup$
3
1
$\begingroup$

Let us show the $4$ subgroups you found are all the subgroups of order $6$ of $S_{4}$. Let $H$ be a subgroup of order $6$. Then $H$ has an element of order $3$. It is a $3$-cycle. If it is $(123)$, we show $H=<(12),(123)>$. $H$ also contains an element $a$ of order $2$. If $a=(12)$, $(13)$, or $(23)$, then we get the subgroup $<(12),(123)>$. If $a=(14)$, we know $(132)\in H$, $(14)(123)(14)=(423)\in H$, $H$ has at least $3$ elements of order $3$, contradiction. Similarly, $a\neq(24),(34)$. Finally, if $a=(12)(34)$, $(12)(34)(123)(12)(34)=(214)\in H$, contradiction. Similarly, $a\neq (13)(24),(14)(23)$. So the $2$ $3$-cycles of $H$ determine all other elements of $H$. Since there are $8$ $3$-cycles in $S_{4}$, there are $4$ subgroups of order $6$.

$\endgroup$

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.