The sole goal of this post will be to prove Lagrange’s theorem. Let be a subgroup of a group . Recall that a *left coset* of is , where . We proved that this is .

**Definition: **Let denote the number of left cosets of in ; this is called *the index of in *.

Suppose . Then let . This is a subgroup of , and we can ask what the cosets of are. Answering this is quite simple, and, indeed, we have nothing but the collection of odd and even integers: and . Thus the index is . More generally, . This should be obvious by a similar observation. The index therefore says “how many copies of form ”.

An important theorem in group theory says that if is a finite group, and if , then is divisible by . More precisely:

**Theorem: **Let be a group, and let . Then , where denotes the *order* of , or the number of elements of .

*Proof. *Recall that each left coset arises as a member of a partition (because it comes from an equivalence class, which in turn gives rise to a unique partition). Let’s digress a little, and show that if , then has the same number of elements as . If we want to show this, it suffices to find maps and which are inverses, i.e., such that is the identity on and is the identity on .

Define by letting . Conversely, define by . Then , and . Thus is a bijection, and and have the same number of elements.

Now, because cosets are all disjoint, each element of is in only one coset. Since the number of left cosets of is , we see that . This is because , and either or , so ; and since for any , we showed that , this becomes where there are copies of , which is given by the number of cosets , which in turn is simply . Consequently, if is finite, then is a positive integer. QED.

Let’s now give an application of Lagrange’s theorem. Let us classify all finite groups up to order five (we choose five because it is the smallest prime number for which there is a number lesser than which is *not* prime). Because any group of prime order is cyclic, if the order of the finite group is , or , then it is cyclic. So suppose is a finite group whose order is . If there’s an element of which has order , then is cyclic (because the order of the subgroup generated by is also ).

Suppose doesn’t contain any elements of order ; by Lagrange’s theorem, this means that contains elements of order . The only element of order is the identity, so every other element must have order . We can draw out a multiplication table for elements of ; drawing this out will be left to you, the reader. Next time, we’ll study, in more detail, orders of elements, etc.

Have fun,

SKD

## One thought on “A Proof of Lagrange’s theorem”