## Lagrange’s Theorem: or how I learned to stop worrying and modulo out by a subgroup.

### June 28, 2010

How could I have been so naive?  How could I have been so myopic?  How is it that I thought I could just wrap up group theory without mentioning Lagrange’s theorem?  How could I let this topic die out not with a bang but with a whimper?

Let us, for old time’s sake, state one more theorem for the group theory primer — and this one’s a biggie!  Remember how division is defined for rational numbers?  $\frac{a}{b}$ sort of means “split $a$ into little piles of size $b$, and $\frac{a}{b}$ is how many piles there are.”  For example, if we have 12 batteries and put them into piles of 3 batteries each, how many piles do we have?  This doesn’t take a rocket scientist.

Well, in fact, groups behave much the same way.  If we have a group $G$ of order (number of elements) 12, and we want to modulo (quotient) out by some normal subgroup $H$ of order 3, we’re left with a bunch of cosets that make up the group $G/H$.  Just how many cosets?  Well, would it surprise you if I said there were…four?

No, of course not.  Let’s state Lagrange’s Theorem.

Theorem (Lagrange): Given a finite group $G$, the order of every subgroup $H$ divides the order of the group $G$, and, furthermore, $|G/H| = \frac{|G|}{|H|}$.  That is, the order of the quotient group is the quotient of the orders.

Let’s observe, first, how adorable this theorem is.  Especially the last part!  It makes group theory seem almost like child’s play!  Let’s note here that I’m not using $H$ as normal, and hence the left and right cosets may not be the same.  I’m going to state without proof that there are the same number of left and right cosets, and so the proof remains unchanged.  If $H$ is not normal, the statement of the theorem changes slightly: we use the notation $[G:H]$, or the index of H in G, defined to be the number of cosets of $H$ in $G$, and so the statement becomes: $[G:H] = \frac{|G|}{|H|}$.  I’m not gonna make a huge deal out of this; just note that $G/H$ is not necessarily defined when $H$ is not normal, and make the substitution of $[G:H]$ for $|G/H|$ in the case that it is not.

Proof. Let’s say $|G| = m$ for some natural number $m$.  Okay, first, let’s show that the cosets $gH$ form a partition of $G$ — that is, let’s show that each element of $G$ is in one and only one coset.  We’ll do this by showing that either two cosets are completely disjoint and share no elements, or if they share one element they must be equal.

Proof that Cosets are either Equal or Disjoint. Alright, suppose $aH$ and $bH$ are not disjoint, so they share some element $x \in aH\cap bH$.  This means that $x = ah_{1} = bh_{2}$, which means that, in particular, $ah_{1} = bh_{2}$ which implies $b = a(h_{1}h_{2}^{-1})\in aH$ and, similarly, $a = b(h_{2}h_{1}^{-1})\in bH$.  This means that for any element $ah\in aH$ we have that $ah = (bh_{2}h_{1}^{-1})h = b(h_{2}h_{1}^{-1}h)\in bH$, and so $aH\subseteq bH$, and, by the same reasoning, we have $bH\subseteq aH$, implying $aH = bH$So two left cosets are either equal or disjoint.

Alright, back.  Now, every element in $G$ can be represented as an element in the form $gh$ for some $g\in G$ and $h\in H$ (why is this true?), and so we have that

$G = g_{1}H \cup g_{2}H\cup \dots \cup g_{n}H$

is a partition of $G$ which contains all elements of $G$ and each element is in exactly one $g_{i}H$.  We need only show that all these cosets have the same size.

Proof that all cosets of $H$ in $G$ have the same size. Alright, let’s make a bijection between them.  Let’s say $f:H\rightarrow bH$ by mapping f(h) = bh.  Clearly this is surjective, since given $bh\in bH$ we simply take $f(h)$.  Is this injective?  If $f(h_{1}) = bh_{1} = bh_{2} = f(h_{2})$, then by multiplying the middle two terms by $b^{-1}$ we show $h_{1} = h_{2}$.  Bam.  Injective.  Therefore, this is a bijection, and so the cosets must have the same size.  In particular, they all have the same size as the coset $H$, which has exactly $|H|$ elements by definition.

Good, okay, back to our proof.  Remember what we were doing?  Go back and read.  We partitioned $G$ into

$G = g_{1}H \cup g_{2}H\cup \dots \cup g_{n}H$

and each of these cosets were disjoint and have the exact same size.  So how many of them are there?  Well, how big is each of them?  $|H|$.  So, we have…

$|G| = m = |H| + |H| + \dots +|H| = n|H|$

or, in other words,

$\frac{m}{|H|} = n$.

Since $n$ is a natural number, this implies that the order of $H$ divides the order of $G$ evenly.

But what is this number of cosets crap?  Partitioning the group $G$ into cosets of $H$ is the same as taking the quotient (or the index if $H$ is not normal).  Therefore, $|G/H|$ has the same order as $\frac{|G|}{|H|}$!  This proves the theorem.  $\Box$