# 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.

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# Proof

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.

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