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

\begin{equation} gH = \{g \Delta h : h \in H \} \end{equation}

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?

\begin{equation} g_1 \sim g_2 \iff g_1 H = g_1 H_2 \end{equation}

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

\begin{equation} |G| = \text {number of equivalence classes} \cdot |H| \end{equation}

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

\begin{equation} |G| = n |H| \end{equation}

That is, we have proven that \(|H|\) divides \(|G|\).