# Converse of Lagrange's Theorem for groups

#### jackmell

I know of only one group, $A_4$ of order 12 which does not have a subgroup with order dividing the group size. In this case, a subgroup of size $6$.

What property of a group causes this? Would I expect to find other examples only in non-abelian groups or are there abelian groups which do not have subgroup orders dividing the group size?

Surely there must be some reason for this anomaly right? Another words, if I looked at say 500 groups of any kind which do not contain subgroups dividing the group size, there would be no defining quality shared by all of these groups which is causing this property?

What other symmetric groups do not contain further subgroups dividing the subgroup size?

I suppose I can always number-crunch it to death and probably find a few . . . 500, not so sure. Ok, here's the task: Find all subgroups of $S_n$ which do not contain subgroups dividing the parent group order. Just how far could I numerically compute that at 2.2 GHz? 20? Any algebraic methods to find them?

Thanks,
Jack

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#### micromass

Well, I can give you some groups for which this doesn't happen: abelian groups and $p$-groups. Those have subgroups of all sizes (as long as Lagrange's theorem is not violated). Also direct products of those and semidirect products.

#### jackmell

Well, I can give you some groups for which this doesn't happen: abelian groups and $p$-groups. Those have subgroups of all sizes (as long as Lagrange's theorem is not violated). Also direct products of those and semidirect products.
Ok thanks. Apparently there is something called CLT-groups (converse to Lagrange Theorem?) which satisfy the converse of Lagrange's Theorem, and likewise non-CLT groups which do not. According to this link:

http://math.stackexchange.com/questions/100933/how-to-prove-that-the-converse-of-lagranges-theorem-is-not-true

there exists non-CLT groups of sizes 12,24,36,48,56,60,72,75,80,84 and 96. We know an example for 12.

Can anyone suggest an example for 24? ( I don't have access to the references cited in the link so can't view the reference)

Here's another reference that looks encouraging (just briefly looked at it):

http://www.emis.de/journals/MPRIA/2004/pa104i1/pdf/104a111.pdf

This reference suggests ALL symmetric groups $S_n$, n>5 are NCLT (non-converse Lagrange Theorem). Also, even more interesting is the quote:

The commutator subgroup G' can indicate if a finite group G is a CLT (Converse Lagrange’s Theorem) group or an NCLT (Non-Converse Lagrange’s Theorem) group.
Now that sounds interesting: how can I use the commutator group to determine if a group is NCLT?

Here it is:

If p is the smallest prime dividing |G| and Pp is a subgroup of G', then G is an NCLT group.
I assume that means, if the commutator group $G'$ has a subgroup of order $p$ and $p$ is the smallest prime dividing $G$, then $G$ is NCLT. Ok, then $|A_4|=12$ and $2|12$ and we know $A_4$ is NCLT. So therefore $A_4$ must have a commutator group having a subgroup of order $2$. That's shouldn't be too hard to verify.

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#### jackmell

I assume that means, if the commutator group $G'$ has a subgroup of order $p$ and $p$ is the smallest prime dividing $G$, then $G$ is NCLT. Ok, then $|A_4|=12$ and $2|12$ and we know $A_4$ is NCLT. So therefore $A_4$ must have a commutator group having a subgroup of order $2$. That's shouldn't be too hard to verify.
Here it is in GAP:
Code:
gap> myagroup:=AlternatingGroup(4);

Alt( [ 1 .. 4 ] )

gap> CommutatorSubgroup(mygroup,mygroup);

Group([ (1,2,3), (1,2,4) ])

gap>
So if I coded that correctly, I assume that means $A_4^{'}=\{c_1 c_2\cdots c_n:\text{c_i is a commutator}\}=\{(1\;2\;3),(1\;2\;4)\}$? I'm not very confident that's correct.

I'm finding GAP hard to use, intractable really but maybe that's because it wasn't written for a Windows platform. Anyone reading this feels like helping me code GAP to find the missing subgroups in as many symmetric groups as GAP can reasonably handle (12 I think)? That is, create a table with column 1 being the group size, and column 2 being the divisor of the group size for which the group does not have a subgroup with that size. I guess though I can get it eventually.

Let's create an OEIS entry: sequence of missing subgroups in the symmetric groups beginning with $S_5$. Might be a problem though because there may be more than one missing subgroup in each symmetric group but I suppose the sequence can list only the "smallest" or "largest" subgroup size missing in each group.

Ok thanks,
Jack

Edit: Wish to correct my interpretation of that code above. The Group construct is actually the group generated by those elements. So Group([(1,2,3),(1,2,4)])); actually means $\big<(1\;2\;3),(1\;2\;4)\big>$and that's a subgroup of order 10. However, note in that subgroup, there is indeed a subgroup of order $2$:
Code:
gap> p5:=AllSubgroups(Group([(1,2,3),(1,2,4)]));

[ Group(()), Group([ (1,2)(3,4) ]), Group([ (1,3)(2,4) ]), Group([ (1,4)(2,3) ]), Group([ (2,4,3) ]),

Group([ (1,2,3) ]), Group([ (1,4,2) ]), Group([ (1,3,4) ]), Group([ (1,2)(3,4), (1,3)(2,4) ]), Group([ (1,2)

(3,4), (1,3)(2,4), (2,4,3) ]) ]

gap> Size(p5);

10

gap> for i in [1..10] do

> Print(Size(p5[i])," ");

> od;

1 2 2 2 3 3 3 3 4 12

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#### jackmell

Got $S_5$:
Code:
gap> mysubs:=AllSubgroups(SymmetricGroup(5));;

gap> Size(mysubs);

156

gap> DivisorsInt(120);

[ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ]

gap> for i in [1..156] do

> Print(Size(mysubs[i])," ");

> od;

1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4

4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 8 8

8 8 8 8 8 8 8 8 8 8 8 8 8 10 10 10 10 10 10 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 20 20 20 20 20 20 24 24 24

24 24 60 120
As you can see in my brilliant GAP code, subgroup sizes of 15, 30, and 40 are not in $S_5$. So then from now on, when the converse of Lagrange's Theorem is mentioned, rather than just state the old "$A_4$ doesn't have a group of size 6" we can now say, "oh, but $S_5$ doesn't have subgroups of 15, 30, and 40 neither". :)

Would be a really tough problem I think: Prove $S_5$ does not contain any sub-groups of order $15$, $30$, and $40$. My approach would be to first study the $A_4$ problem which I kinda' can prove, then scale-up to maybe $A_5$ if it even has missing groups, then hopefully succeed at $S_5$.

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#### jackmell

I got the final code to compute the missing groups. The code below took 9 seconds to find (all?) $11300$ subgroups for $S_7$. The final list of 29 are the group sizes missing in $S_7$ assuming GAP is correct.
Code:
gap> notlist:=;

[ 0 ]

gap> gIndex:=7;

7

gap> thesubs:=AllSubgroups(SymmetricGroup(gIndex));;

gap> subsize:=Size(thesubs);

11300

gap> theorders:=[1..subsize];

[ 1 .. 11300 ]

gap> thediv:=DivisorsInt(Factorial(gIndex));

[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 28, 30, 35, 36, 40, 42, 45, 48, 56, 60, 63, 70, 72,

80, 84, 90, 105, 112, 120, 126, 140, 144, 168, 180, 210, 240, 252, 280, 315, 336, 360, 420, 504, 560, 630, 720,

840, 1008, 1260, 1680, 2520, 5040 ]

gap> divsize:=Size(thediv);

60

gap> for i in [1..subsize] do

> theorders[i]:=Size(thesubs[i]);

> od;

gap> for i in [1..divsize] do

> if(thediv[i] in theorders)=false then Add(notlist,thediv[i]);

> fi;

> od;

gap> Remove(notlist,1);

0

gap> List(notlist);

[ 15, 28, 30, 35, 45, 56, 63, 70, 80, 84, 90, 105, 112, 126, 140, 180, 210, 252, 280, 315, 336, 420, 504, 560, 630,

840, 1008, 1260, 1680 ]

gap>

#### FireGarden

This question is not quite what I expected by the title. I'm not sure your question will have a simple (i.e., human-accessible) answer, since groups are incredibly general overall (i.e., require relatively little structure). But supposing you're interested in an actual converse of Lagrange (i.e., given a group, does it have subgroups for each factor of its order?) there is a partial answer given by the Sylow theorems.

The first theorem states that if $p^n$ divides the order of G, then G has a subgroup of that order. As you notice for A_4, it doesn't have a subgroup of order 6 - well, order 6 is the only possible subgroup it could've been missing! The second and third theorems are just a bit more about the structure; the second says all such groups (they're called Sylow p-subgroups) of a given order are conjugate to each other, and the third says (amongst other things) that the number of such groups in G is congruent to 1 mod p.

So all prime-powers dividing the order of a group correspond to subgroups - sometimes many. Thus in the special case where G has a prime-power order, it has subgroups for every divisor.