# Understanding Lagrange's Theorem (Group Theory)

I am beginning with Abstract Algebra and I'm trying to understand Lagrange's Theorem. The theorem reads

For any finite group $G$, the order of every subgroup $H$ of $G$ should divide the order of $G$.

It seems simple and I've used it to solve some exercices but I believe I'm missing the essence of it. Is there an example that can help me understand it better, or visualize it? Is there a geometric interpretation?

-
There is a local-representation theoretical version: if $H$ is a subgroup of $G$, then every projective module of $H$ induces a projective module of $G$. Hope you like it. :D –  awllower Mar 21 '13 at 12:01
If you like to think visually, this books.google.com/… might pique your interest (not only about Lagrange's theorem, but in general)! –  Stahl Mar 21 '13 at 12:06
The proof works by considering the cosets of $H$ in $G$, what we see is that the cosets partition $G$, so then $G$ is in some sense covered by sets of size $H$ which do not overlap. In terms of visualisation, draw a big circle, call it $G$, then you can partition your circle into equal sections of size $|H|$, for any subgroup $H$. –  user27182 Mar 21 '13 at 12:08
What would qualify as "the essence" of this? It is a very straightforward theorem. It has no hidden details or mysterious manipulations. –  Pedro Tamaroff Mar 21 '13 at 13:03
@Stahl This book is amazing, thank you! It turns out the author has also developed an open source application called GroupExplorer that helps visualize groups, homomorphisms, subgroup lattices, and more, which I find it incredibly useful in my case. –  gpo Mar 23 '13 at 21:12

The left-multiplication map $x\mapsto ax$ is bijective on $G$; its injectivity follows from the cancellative property of the group's operation, $ah=ag\iff h=g$, and overall bijectivity is a consequence of the fact that it has an inverse map, $x\mapsto a^{-1}x$. I like to view a subgroup $H\le G$ as a "puck" and the overgroup $G$ as a "air-hockey table" on which $H$ resides, and to move $H$ around we apply left multiplication by various elements. If you left-multiply by an element $a\in H$, you have not moved the puck at all since $a\in H\iff H=aH$.

Every element $g\in G$ is in some coset, or left translate, of $H$ - in particular, $g=ge\in gH$ since we know that $e\in H$. Thus, the collection of all translates (possible places for the puck to be positioned in) of $H$ "cover" the entire air-hockey table. It remains, then, to investigate the nature of the overlaps between positions, i.e. the intersections of distinct cosets. Here is the proof that cosets that overlap nontrivially must in fact be identical, put into visual form:

$\hskip 0.6in$

This means the cosets of $H$ partition the group $G$. As left multiplication is bijective, every coset is the same size, so each "looks" the same from the viewpoint of cardinality. Continuing with the idea of an air hockey table, this tells us the puck positions tile it, so we have something like:

$\hskip 1.2in$

The most fundamentally basic meaning imputed to multiplication of natural numbers is the following: if Alice has $n$ bags each containing $m$ apples, then she has $n\times m$ apples total. Similarly, our group $G$ is covered by some number $[G:H]$ of disjoint cosets, each containing $|H|$ elements, so $|G|=[G:H]\times|H|$. Note that this is even true on the level of arbitrary infinite cardinals. Thus, $|H|$ is a divisor of the order $|G|$: Lagrange's theorem.

The converse is not globally true: not every divisor $d$ of $n=|G|$ corresponds to a subgroup $H\le G$ of size $|H|=d$. Sylow theory however yields a local version of a converse: for every prime power $q=p^r$ that is a divisor $q\mid n$, there is a $p$-subgroup $H$ of size $|H|=q$.

-
Thank you so much for the time and energy you put in writing such a comprehensive and detailed answer! It really helped me a lot and makes things a lot clearer. I actually understand the proof our professor gave now! –  gpo Mar 23 '13 at 19:46

You already have several great explanations, but you asked also for a way to visualise this. One way to visualise Lagrange's Theorem is to draw the Cayley table of (smallish) groups with colour highlighting.

Here is the Cayley table of a dicyclic group of order $16$ with the cosets of its centre of order $2$ highlighted. The subgroup itself consists of the elements $\{ e, a \}$ (where $e$ is the identity), and is shown in red. The other cosets appear with different colours. Because the subgroup, in this case, is normal, the table is highly regular.

# The Maple code to produce this is:
> with( GroupTheory ):
> G := DicyclicGroup( 4 ):
> H := Centre( G  ):
> DrawCayleyTable( G, cosets = H );


Here is another example, in which the subgroup is not normal. It is the Sylow $2$-subgroup of the symmetric group $S_{4}$. Since the order of $S_{4}$ is equal to $24$, the Sylow $2$-subgroup has order $8$ (and index $3$), and each coset has the same size $8$ as the subgroup. These facts are clear from the picture.

Nevertheless, the block structure of the cosets is still quite visible.

# Maple code for the second example:
> G := Symm( 4 ):
> DrawCayleyTable( G, cosets = SylowSubgroup( 2, G ) );


In each case, it is visually clear that the cosets of the subgroup form a "regular" partition of the group elements. The sizes of the blocks forming each coset are all identical, so they form a regular tiling of the Cayley table for the entire group.

Hope this helps!

-
Those are some very nice visualisations, thank you. I had no idea you could do this in Maple! –  gpo Mar 23 '13 at 20:04

I think the key idea is that groups admit "translations", that is transformation of the form $\left\{ \begin{array}{ccc} G & \to & G \\ g & \mapsto & h \cdot g \end{array} \right.$ with $h \in G$. Moreover, if $H$ is a subgroup of $G$, then the images of $H$ by two such translations are either equal or disjoint; therefore, you can cover $G$ by translating the subgroup $H$ to get a partition of $G$ into $n$ disjoint copies of $H$. You deduce that $|G|=n|H|$.

-
Any equivalence relation on a set $\,S\,$ induces a partition on $\,S\,$ (and vice versa). This is a special case where the $\,n\,$ equivalence classes have the same size $\,k,\,$ so $\,|S| = nk,\:$ so $\,k\,$ divides $|S|.\,$ The result depends crucially on this key fact that the classes have the same size (which you do not mention). –  Math Gems Mar 21 '13 at 14:38
@MathGems: I didn't give a proof but a way to understand Lagrange's theorem, and intuitively, a translation does not change the size. Moreover, implicitely a copy of $H$ has the same size than $H$. –  Seirios Mar 21 '13 at 14:45
@Serios The word "translation" is used in many ways in mathematics, not all of which are set-theoretical bijections. In any case, from a pedagogical standpoint, it is always beneficial to bring to the fore those facts which play key roles. Hence my prior comment. –  Math Gems Mar 21 '13 at 14:49

The main thing is that there is a natural one-to-one correspondence between any two (say, left) cosets of the subgroup $H$, namely the mapping $$s\mapsto yx^{-1}s$$ takes the coset $xH=\{xh\,\mid h\in H\}$ to $yH$, and its inverse is $s\mapsto xy^{-1}s$ (because $xy^{-1}yx^{-1}s=s$ for all $s$).

So, the size of each coset is the same ( $=|H|$), so, if $|G|$ is finite, $|H|$ must divide it. Stahl commented a nice link for visualizing this fact..

-

For example If order of G is 12 then we can find only subgroups of order 1,2,3,4,6 and 12 which are divisiors of 12

It means that we can not find a subgroup of other order except above orders

-