# Thread:

1. ## Deduce Lagrange's Theorem

Let H be a group acting on a set A. Prove that the relation ~ on A defined by a~b iff $a=hb$ for some h in H is an equivalence relation.
I have already shown this.

Let H be a subgroup of the finite group G and let H act on G by left multiplication. Let x exist in G and let O be the orbit of x under the action of H. Prove that the map $H\to O$ defined by $h\mapsto hx$ is a bijection.
I have already shown this too.

From these two statements, deduce Lagrange's Theorem: if G is a finite group and $H\leq G$, then $|H|| |G|$.

I understand Lagrange's Theorem. I need an explanation on how I can deduce his theorem from the above.

2. ## Re: Deduce Lagrange's Theorem

Originally Posted by dwsmith
I understand Lagrange's Theorem. I need an explanation on how I can deduce his theorem from the above.
Hint All the equivalence classes have the same cardinality.

3. ## Re: Deduce Lagrange's Theorem

Originally Posted by FernandoRevilla
Hint All the equivalence classes have the same cardinality.
I know that since each one is a coset but I still don't understand.

4. ## Re: Deduce Lagrange's Theorem

if every coset has the same size, then they all have the same size as the coset containing the identity, which is H.

so the coset containing the identity, H, has cardinality |H|, as does every coset Hg.

so, summing over our partition of G:

|G| = |H| + |Hg1| + |Hg2| +....+ |Hgk|, where k is the number of distinct cosets. and....?

5. ## Re: Deduce Lagrange's Theorem

Originally Posted by dwsmith
I know that since each one is a coset but I still don't understand.
The set $G$ has the form $G=\cup_{i}C_i$ and the family of cosets $\mathcal{C}=\{C_i\}$ is pairwise disjoint with $|C_i|=|H|$. So, $|G|=|\mathcal{C}||H|$ .

6. ## Re: Deduce Lagrange's Theorem

and the pairwise-disjointness arises because being in the same coset is an equivalence relation (what you proved at the beginning), and equivalence classes partition the set they are on.

(equivalence classes are "quotient sets").
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