Joseph Louis-Lagrange
1. Subgroups
Last time we talked about the symmetry group of a triangle, \(D_3\), and found it had six elements: \(e\), \(r\), \(r^2\), \(f\), \(rf\), \(r^2f\), subject to \(r^3 = e\), \(f^2 = e\) and \(fr = r^2f\). Now the obvious next question - inside that group, are there smaller groups? Not just smaller collections of elements, but subsets that are themselves closed, that contain the identity, and that contain their own inverses. A group hiding inside a group. Obviously, this is math, so the answer is yes, and there’s actually a very cool way that they are distributed.
1.1 What counts as a subgroup
A subset \(H \subseteq G\) is a subgroup if it forms a group under the same operation. Basically it means three things: \(e \in H\), the product of any two elements of \(H\) stays in \(H\), and every element of \(H\) has its inverse in \(H\). Associativity follows, since it held in \(G\) and \(H\) isn't doing anything different. Every group has two subgroups at least: \({e}\) on its own, and \(G\) itself. Those are the trivial ones.
1.2 Hunting them down in \(D_3\)
Let's just look. Start with the rotations:
\[ H_1 = { e, r, r^2 } \tag{1} \]
Closed? \(r \cdot r = r^2\), \(r \cdot r^2 = r^3 = e\), \(r^2 \cdot r^2 = r^4 = r\). Everything stays inside. Inverses? \(r^{-1} = r^2\) and \((r^2)^{-1} = r\), both present. So yes, a subgroup of order 3. Next, take a single reflection with the identity:
\[ H_2 = { e, f } \tag{2} \]
Since \(f^2 = e\), that's closed and \(f\) is its own inverse. A subgroup of order 2. The same works for the other two reflections:
\[ (rf)(rf) = r(fr)f = r(r^2f)f = r^3f^2 = e \tag{3} \]
So \(rf\) squares to the identity too, and \({e, rf}\) is a subgroup. Likewise \({e, r^2f}\). Here’s a failure. Take \({e, r, f}\) and see whether it closes:
\[ r \cdot f = rf \tag{4} \]
which isn't in the set. Throw \(rf\) in, and now you need \(r \cdot rf = r^2f\) as well, and \(f \cdot r = r^2f\), and before long you've got all six elements back. Mixing a rotation of order 3 with a reflection generates the whole group. So the complete list is:
\[ {e}, \quad {e, f}, \quad {e, rf}, \quad {e, r^2f}, \quad {e, r, r^2}, \quad D_3 \tag{5} \]
with orders \(1, 2, 2, 2, 3, 6\). Anything interesting about those numbers? Don’t worry if you didn’t notice, I didn’t either. Turns out they’re the factors of 6, which makes you wonder if there are any similarities between numbers and groups.
1.3 Cosets
Anyway, here's the construction that explains it. Take a subgroup \(H\) and an element \(g \in G\), and form the set of everything in \(H\) multiplied on the left by \(g\):
\[ gH = { gh : h \in H } \tag{6} \]
This is a left coset. It's usually not a subgroup (if \(g \notin H\) it won't even contain the identity) but that's fine, we only want it as a counting device. Do it with \(H = {e, f}\):
\[ eH = { e, f }, \qquad rH = { r, rf }, \qquad r^2H = { r^2, r^2f } \tag{7} \]
Three sets, two elements each, six elements total, and every element of \(D_3\) appears exactly once. Turns out you don't get anything new by picking other elements. Try \(g = f\):
\[ fH = { fe, ff } = { f, e } = eH \tag{8} \]
Same coset, just written differently. Every choice of \(g\) lands on one of the three we already have.
1.4 Why they are equal
Every coset is the same size as \(H\). Map \(H \to gH\) by sending \(h \mapsto gh\). It hits everything in \(gH\) by definition. And it's injective, because if \(gh_1 = gh_2\) then multiplying by \(g^{-1}\) on the left gives \(h_1 = h_2\). A bijection, so \(|gH| = |H|\). Two cosets are either identical or disjoint. Suppose \(aH\) and \(bH\) share an element, so \(ah_1 = bh_2\) for some \(h_1, h_2 \in H\). Rearranging, \(a = bh_2h_1^{-1}\). Then for any \(h \in H\),
\[ ah = b h_2 h_1^{-1} h \tag{9} \]
and \(h_2h_1^{-1}h\) is a product of elements of \(H\), so it's in \(H\). That puts \(ah\) in \(bH\). Since \(h\) was arbitrary, \(aH \subseteq bH\), and running the identical argument the other way gives \(bH \subseteq aH\). So they're the same set. Finally, every \(g \in G\) lies in some coset, namely \(gH\), since \(g = ge\). So the cosets cover \(G\) completely.
1.5 Lagrange's theorem
Put those together; the cosets of \(H\) partition \(G\) into disjoint pieces, and every piece has exactly \(|H|\) elements. If there are \(k\) pieces, then
\[ |G| = k \cdot |H| \tag{10} \]
and therefore \(|H|\) divides \(|G|\). That's Lagrange's theorem, and the number \(k\) of cosets is called the index of \(H\) in \(G\). It explains the list above completely. The subgroups of \(D_3\) have orders 1, 2, 3, 6 because those are the divisors of 6, and there's no order-4 or 5 subgroups because 4 and 5 don’t divide 6. We also are given a useful corollary. For any element \(g\), the powers \({e, g, g^2, \dots}\) form a subgroup, and its size is the order of \(g\), meaning the smallest \(n\) with \(g^n = e\). So the order of every element divides the order of the group. In \(D_3\): the reflections have order 2, the rotations order 3, the identity order 1, and \(2, 3, 1\) all divide 6. So if \(|G|\) is prime, the only divisors are 1 and \(|G|\), so there are no subgroups except the trivial ones. Pick any element other than \(e\), and the subgroup it generates has to be everything. Every group of prime order is generated by a single element, and there's basically only one such group for each prime. Cool!
1.6 One thing that doesn't work
We built \(gH\) by multiplying on the left. We could just as well have multiplied on the right, forming \(Hg\). Does it matter? Take \(H = {e, f}\) again and compare:
\[ rH = { r, rf }, \qquad Hr = { r, fr } = { r, r^2f } \tag{11} \]
Different sets. So left and right cosets are not the same thing. But now try it with the rotation subgroup \(H_1 = {e, r, r^2}\):
\[ fH_1 = { f, r^2f, rf }, \qquad H_1f = { f, rf, r^2f } \tag{12} \]
Identical. So some subgroups have left cosets matching their right cosets and some don't, and \(D_3\) contains an example of each. In fact, that’s exactly the condition needed to turn the set of cosets into a group in its own right.