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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.

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3 Answers 3

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.

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Why each podgroup contains $4$ such cycles? –  Thomas Sep 4 '13 at 18:33
1  
@Tomek: Every nonidentity element in a group of order $5$ has order $5$. There are $4$ non-identity elements in a group of order $5$. –  Jared Sep 4 '13 at 19:01
    
I don't understand. For example I can take $(1,2,3,4,5),(2,3,4,5,6),(3,4,5,6,7),(1,2,7,4,5),(2,3,4,5,6),(3,4,1,6,7)$ so I have $6$ non-identity elements... –  Thomas Sep 4 '13 at 19:19
    
@Tomek Those are not members of the same 5-element subgroup! The point is that for each subgroup of order 5, there are exactly four elements of order 5 (the four non-identity members of that subgroup). –  Steven Stadnicki Sep 4 '13 at 19:38
    
@Tomek: The elements you've listed do not form a subgroup. Consider the subgroup with elements $e,(1,2,3,4,5),(1,3,5,2,4),(1,4,2,5,3),$ and $(1,5,4,3,2)$, and notice that there are four non-identity elements, and they all have order $5$. Try listing out the elements of the subgroup generated by $(3,4,5,6,7)$. –  Jared Sep 4 '13 at 19:39

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
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+ 1 $\ddot\smile!$ –  amWhy Oct 23 '13 at 1:18

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

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

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This is a problem, that i don't know how many $5$ cycles there are, and which of them generete the same subgroup. Of course, we have ${7 \choose 5} = 21 $ $5$-cycles. The problem is, that I don't know which of them are non-identity and which of them generate non-identity subgroup. –  Thomas Sep 5 '13 at 17:12
    
Well Tomek, one has to know something to attack this kind of problems in medium undergraduate level... –  DonAntonio Sep 5 '13 at 18:11

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