Subgroups of $S_4$ of order 6

Question

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)}

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 $S_4$. — Gerry Myerson, Jan 28 '16 at 11:32

score 1 · Answer 1 · answered Jan 28 '16 at 11:36

1

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 $S_4$.

answered Jan 28 '16 at 11:36

lhf

221,500

See also http://groupprops.subwiki.org/wiki/S3_in_S4 and http://math.stackexchange.com/questions/379841/how-to-enumerate-subgroups-of-each-order-of-s-4-by-hand. – lhf Jan 28 '16 at 11:53
why are these subgroups isomorphic to $S_3$? – Giulia B. Jan 28 '16 at 11:58
@GiuliaB., because they'd cyclic otherwise, as mentioned. – lhf Jan 28 '16 at 12:12

score 1 · Answer 2 · answered Jan 29 '16 at 03:49

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 $6$.

Subgroups of $S_4$ of order 6

2 Answers2