How many $5$-element subgroups does $S_7$ have?

Question

How many $5$-element subgroups are there in $S_7$, the group of permutations on $7$ elements?

Let $H$ be a $5$-element subgroup of $S_7$. We have $\mbox{ord}(H) = 5$ and $5\mid 7!$. But I don't have any idea how can I find 5-element subgroups.

Answer 1

Hint:

A subgroup of $S_7$ of order $5$ must be cyclic (all groups of prime order are cyclic), and therefore is generated by an element of order $5$. The only elements of order $5$ in $S_7$ are $5$-cycles (why?), but each subgroup contains $4$ such cycles. This reduces the problem to one of a combinatorial flavor.

Answer 2

Hints (completing and/or adding to some other comments or answers):

• Each group of order $\;5\;$ in $\;S_7\;$ is generated by a $\,5-$cycle (this is false in $\,S_n\;,\;\;n\ge10\;$ . Can you see why?)

• For each given subset of five elements in $\;\{1,2,3,4,5,6,7\}\;$ we get a generator of a subgroup of order $\;5\;$ in $\;S_7\;$ and all of them yield different subgroup of order $\;5\;$ in $\;S_7\;$ .

• (kind of) Another approach: How many $\;5\;$ cycles are there in $\,S_7\;$ ? How many of them generate the same cyclic subgroup? For example

$$\langle (12345)\rangle =\langle(13524)\rangle=\langle(14253)\rangle =\ldots$$

Answer 3

Here are some codes inspired by A. Konovalov's leading suggestions for doing the problem via GAP.

gap> G:=SymmetricGroup(7);
gap> ccs:=ConjugacyClassesSubgroups(G);;
gap> c5:=Filtered(ccs, c -> Size(Representative(c))=5);;
gap> Length(c5);
                                       1
gap> c5;
                      [ Group( [ (3,4,5,6,7) ] )^G ]
gap> Size(c5[1]);
                                      126

The conjugacy class of a 5-cycle contains 504 elements which is 4 times more:

gap> (3,4,5,6,7)^G;
gap> Size(last);
                                      504
gap> 504/126;
                                       4