# A calculus of the absurd

### Chapter 26 Abstract algebra

#### 26.1 Groups

##### 26.1.1 Lagrange’s theorem
• Theorem 26.1.1 Let $$(G, \Delta )$$ be a finite group, and $$(H, \Delta )$$ a subgroup of $$G$$, then $$|H|$$ divides $$|G|$$.

This is a really powerful theorem (or so I’m told 142142 Well, it certainly has its uses.).

Let’s start by considering the set $$gH$$ for $$g \in G$$. Wait, that’s not a set! (Well not written like that at least, it’s not)! We define this as

$$gH = \{g \Delta h : h \in H \}$$

Here’s one of those magic tricks we can pull as though straight from a hat; we can define an equivalence relation $$\sim$$ on these sets! How?

$$g_1 \sim g_2 \iff g_1 H = g_1 H_2$$

This is an equivalence relation (TODO: proof) on the set $$G$$, and very importantly, equivalence relations form a partition on the set $$G$$ (proof in Section TODO: write section).

Now the claim is that for any $$g$$, we have that $$|gH| = |H|$$ and thus that

$$|G| = \text {number of equivalence classes} \cdot |H|$$

Therefore, we just need a bijection $$\phi : H \to gH$$, which we can obtain by defining $$h \mapsto gh$$. Then we just need to prove that $$\phi$$ is a bijection!

• 1. Injective. Assume $$\phi (a) = \phi (b)$$, then $$ga = gb$$, thus $$g^{-1}(ga) = g^{-1}(gb)$$ and therefore $$(g^{-1}g)a = (g^{-1}g)b$$, from which we have that $$a = b$$, which means that $$\phi$$ is injective.

• 2. Surjective. Fix $$k \in gH$$, then $$k = gh$$ for some $$h$$, thus we have that $$k = \phi (h)$$, so $$\phi$$ is surjective.

Therefore, we have some integer $$n$$ (equal to the number of equivalence classes) such that

$$|G| = n |H|$$

That is, we have proven that $$|H|$$ divides $$|G|$$.