# Lagrange’s Theorem

**Lagrange’s Theorem**: Lagrange’s Theorem states a very important relation between the orders of a finite group and any subgroup.

**Theorem 3.15 (Lagrange’s Theorem)** Let H be a subgroup of a finite group G. Then the order of H divides the order of G.

**Proof **We already know from the last section that the group G is partitioned into the right cosets of H. We show that every right coset Hg contains the same number of elements as H.

To prove this, we construct a bijection f from H to H_{g}. The bijection is defined in the obvious way: Φ maps h to hg.

• Φ is one-to-one: suppose that Φ(h_{1}) = Φ)h_{2}, that is, h_{1}g = h_{2}g. Cancelling the g (by the cancellation law, or by multiplying by g^{−1}), we get h_{1} = h_{2}.

• Φ is onto: by definition, every element in the coset Hg has the form hg for some h ∈ H, that is, it is Φ(h).

So f is a bijection, and |H_{g}| = |H|.

Now, if m is the number of right cosets of H in G, then m|H| = |G|, so |H| divides |G|.

Remark We see that |G|/|H| is the number of right cosets of H in G. This number is called the index of H in G.

We could have used left cosets instead, and we see that |G|/|H| is also the number of left cosets. So these numbers are the same. In fact, there is another reason for this:

Exercise Show that the set of all inverses of the elements in the right coset Hg form the left coset g−1H. So there is a bijection between the set of right cosets and the set of left cosets of H.

In the example in the preceding section, we had a group S3 with a subgroup having three right cosets and three left cosets; that is, a subgroup with index 3.

**Corollary 3.16** let g be an element of the finite group G. Then the order of g divides the order of G.

**Proof **Remember, first, that the word “order” here has two quite different meanings: the order of a group is the number of elements it has; while the order of an element is the smallest n such that g^{n} = 1.

However, we also saw that if the element g has order m, then the set {1,g,g2, . . . ,g^{m−1}} is a cyclic subgroup of G having order m. So, by Lagrange’s Theorem, m divides the order of G.

**Example** Let G = S3. Then the order of G is 6. The element (1)(2,3) has order 2, while the element (1,3,2) has order 3.