Lagrange theorem on subgroup size: Intuitive version

Given a finite group \(G\), it may have many sub­groups. So far, we know al­most noth­ing about those sub­groups; it would be great if we had some way of re­strict­ing them.

An ex­am­ple of such a re­stric­tion, which we do already know, is that a sub­group \(H\) of \(G\) has to have size less than or equal to the size of \(G\) it­self. This is be­cause \(H\) is con­tained in \(G\), and if the set \(X\) is con­tained in the set \(Y\) then the size of \(X\) is less than or equal to the size of \(Y\). (This would have to be true for any rea­son­able defi­ni­tion of “size”; the usual defi­ni­tion cer­tainly has this prop­erty.)

La­grange’s The­o­rem gives us a much more pow­er­ful re­stric­tion: not only is the size \(|H|\) of \(H\) less than or equal to \(|G|\), but in fact \(|H|\) di­vides \(|G|\).

A pri­ori, all we know about the sub­groups of the cyclic group \(C_6\) of or­der \(6\) is that they are of or­der \(1, 2, 3, 4, 5\) or \(6\).

La­grange’s The­o­rem tells us that they can only be of or­der \(1, 2, 3\) or \(6\): there are no sub­groups of or­der \(4\) or \(5\). La­grange tells us noth­ing about whether there are sub­groups of size \(1,2,3\) or \(6\): only that if we are given a sub­group, then it is of one of those sizes.

In fact, as an aside, there are in­deed sub­groups of sizes \(1,2,3,6\):

  • the sub­group con­tain­ing only the iden­tity is of or­der \(1\)

  • the “im­proper” sub­group \(C_6\) is of or­der \(6\)

  • sub­groups of size \(2\) and \(3\) are guaran­teed by Cauchy’s the­o­rem.



In or­der to show that \(|H|\) di­vides \(|G|\), we would be done if we could di­vide the el­e­ments of \(G\) up into sep­a­rate buck­ets of size \(|H|\).

There is a fairly ob­vi­ous place to start: we already have one bucket of size \(|H|\), namely \(H\) it­self (which con­sists of some el­e­ments of \(G\)). Can we per­haps use this to cre­ate more buck­ets of size \(|H|\)?

For mo­ti­va­tion: if we think of \(H\) as be­ing a col­lec­tion of sym­me­tries (which we can do, by Cayley’s The­o­rem which states that all groups may be viewed as col­lec­tions of sym­me­tries), then we can cre­ate more sym­me­tries by “tack­ing on el­e­ments of \(G\)”.

For­mally, let \(g\) be an el­e­ment of \(G\), and con­sider \(gH = \{ g h : h \in H \}\).

Ex­er­cise: ev­ery el­e­ment of \(G\) does have one of these buck­ets \(gH\) in which it lies.

The el­e­ment \(g\) of \(G\) is con­tained in the bucket \(gH\), be­cause the iden­tity \(e\) is con­tained in \(H\) and so \(ge\) is in \(gH\); but \(ge = g\).

Ex­er­cise: \(gH\) is a set of size \(|H|\). noteMore for­mally put, left cosets are all in bi­jec­tion.

In or­der to show that \(gH\) has size \(|H|\), it is enough to match up the el­e­ments of \(gH\) bi­jec­tively with the el­e­ments of \(|H|\).

We can do this with the func­tion \(H \to gH\) tak­ing \(h \in H\) and pro­duc­ing \(gh\). This has an in­verse: the func­tion \(gH \to H\) which is given by pre-mul­ti­ply­ing by \(g^{-1}\), so that \(gx \mapsto g^{-1} g x = x\). <div><div>

Now, are all these buck­ets sep­a­rate? Do any of them over­lap?

Ex­er­cise: if \(x \in rH\) and \(x \in sH\) then \(rH = sH\). That is, if any two buck­ets in­ter­sect then they are the same bucket. noteMore for­mally put, Left cosets par­ti­tion the par­ent group.

Sup­pose \(x \in rH\) and \(x \in sH\).

Then \(x = r h_1\) and \(x = s h_2\), some \(h_1, h_2 \in H\).

That is, \(r h_1 = s h_2\), so \(s^{-1} r h_1 = h_2\). So \(s^{-1} r = h_2 h_1^{-1}\), so \(s^{-1} r\) is in \(H\) by clo­sure of \(H\).

By tak­ing in­verses, \(r^{-1} s\) is in \(H\).

But that means \(\{ s h : h \in H \}\) and \(\{ r h : h \in H\}\) are equal. In­deed, we show that each is con­tained in the other.

  • if \(a\) is in the right-hand side, then \(a = rh\) for some \(h\). Then \(s^{-1} a = s^{-1} r h\); but \(s^{-1} r\) is in \(H\), so \(s^{-1} r h\) is in \(H\), and so \(s^{-1} a\) is in \(H\). There­fore \(a \in s H\), so \(a\) is in the left-hand side.

  • if \(a\) is in the left-hand side, then \(a = sh\) for some \(h\). Then \(r^{-1} a = r^{-1} s h\); but \(r^{-1} s\) is in \(H\), so \(r^{-1} s h\) is in \(H\), and so \(r^{-1} a\) is in \(H\). There­fore \(a \in rH\), so \(a\) is in the right-hand side. <div><div>

We have shown that the “cosets\(gH\) are all com­pletely dis­joint and are all the same size, and that ev­ery el­e­ment lies in a bucket; this com­pletes the proof.