Evariste Galois
1. Homomorphisms and Quotients
Inside \(D_3\), the subgroup \({e, f}\) has left cosets that differ from its right cosets, while the rotation subgroup \({e, r, r^2}\) has them agreeing exactly. Weird. What’s this about? This difference is arguably the single most important distinction in elementary group theory, because it's the condition under which you can take a group, collapse a subgroup down to a point, and still have a group left.
1.1 Normal subgroups
A subgroup \(N \subseteq G\) is normal if its left and right cosets are equal:
\[ gN = Ng \qquad \text{for every } g \in G \tag{1} \]
It's usually easier to test the rearranged version. Multiply on the right by \(g^{-1}\) and the condition becomes
\[ gNg^{-1} = N \tag{2} \]
which says \(N\) is unchanged by conjugation. So take anything in \(N\), conjugate it by anything in \(G\), and check you are still in \(N\). Let's try it on both subgroups from last time. Start with the rotations, \(N = {e, r, r^2}\), and conjugate \(r\) by \(f\). Since \(f^{-1} = f\), and using \(fr = r^2f\) from post 1:
\[ frf^{-1} = frf = (r^2f)f = r^2f^2 = r^2 \tag{3} \]
which is back in \(N\). Conjugating by rotations obviously does nothing, so \(N\) is normal. Now the reflections, \(H = {e, f}\), conjugating \(f\) by \(r\). We need \(fr^2\) first:
\[ fr^2 = (fr)r = (r^2f)r = r^2(fr) = r^2(r^2f) = r^4 f = rf \tag{4} \]
so that
\[ rfr^{-1} = rfr^2 = r(fr^2) = r(rf) = r^2f \tag{5} \]
and \(r^2f\) is not in \(H\). So \(H\) isn't normal.
1.2 Why normality is the right condition
Here's what we actually want to do. Given a subgroup \(N\), the cosets partition \(G\) into equal pieces. It would be lovely if those pieces formed a group in their own right, with multiplication defined the obvious way:
\[ (aN)(bN) = abN \tag{6} \]
The problem is that a coset has many names. The set \(aN\) is equally well written \(a'N\) for any \(a' \in aN\), and similarly for \(b\). So before this definition means anything, we have to check that the answer doesn't depend on which names we picked. Take \(a' = ah_1\) and \(b' = bh_2\) with \(h_1, h_2 \in N\). Their product is
\[ a'b' = ah_1bh_2 \tag{7} \]
and we need this in \(abN\). The issue is the \(h_1\) stuck in the middle, between \(a\) and \(b\). We'd like to slide it past \(b\). Normality lets us do that. Since \(b^{-1}h_1b \in N\), call it \(h_3\), which rearranges to \(h_1b = bh_3\). Substituting,
\[ a'b' = a(bh_3)h_2 = ab(h_3h_2) \tag{8} \]
and \(h_3h_2 \in N\), so \(a'b' \in abN\). The product is well defined. Without normality the whole thing collapses, because you can't move elements of \(N\) past elements of \(G\), and the "product" of two cosets depends on which representatives you have written down. When it does work, the resulting group is the quotient group \(G/N\), whose elements are the cosets.
1.3 A quotient we can see
Take \(G = D_3\) and \(N = {e, r, r^2}\). There are two cosets:
\[ N = { e, r, r^2 }, \qquad fN = { f, rf, r^2f } \tag{9} \]
the rotations and the reflections. Multiply them:
\[ N \cdot N = N, \qquad N \cdot fN = fN, \qquad fN \cdot fN = f^2 N = N \tag{10} \]
A group with two elements where one is the identity and the other squares to it. So
\[ D_3 / N \cong \mathbb{Z}_2 \tag{11} \]
And notice what the quotient has done. It's thrown away every detail about which rotation or which reflection, and kept only the one bit of information that stays behind, i.e. rotation or reflection. Two reflections compose to a rotation, which is why the quotient is \(\mathbb{Z}_2\) and not something else. That's the general idea. A quotient is a “controlled loss of information”, and normality is the condition that the remaining information is still a group.
1.4 Homomorphisms
Now go from the other direction. A homomorphism is a map \(\varphi: G \to G'\) that respects the group operation:
\[ \varphi(ab) = \varphi(a)\varphi(b) \tag{12} \]
Multiply first and then map, or map first and then multiply, and you get the same answer. A homomorphism that's also a bijection is an isomorphism, which is what we meant in post 1 when we said \(D_3\) and \(S_3\) were the same group. Two small consequences fall straight out. Setting \(a = b = e\) gives \(\varphi(e) = \varphi(e)\varphi(e)\), and cancelling gives
\[ \varphi(e) = e' \tag{13} \]
and then \(\varphi(g)\varphi(g^{-1}) = \varphi(e) = e'\) gives
\[ \varphi(g^{-1}) = \varphi(g)^{-1} \tag{14} \]
So identities go to identities and inverses to inverses, without either being assumed. Here's a homomorphism on \(D_3\). Send every rotation to \(+1\) and every reflection to \(-1\):
\[ \varphi: D_3 \to { +1, -1 } \tag{15} \]
Is it one? Two reflections compose to a rotation, and \((-1)(-1) = +1\). A reflection times a rotation is a reflection, and \((-1)(+1) = -1\). Great! It works.
1.5 Kernels are normal
The kernel of \(\varphi\) is everything sent to the identity:
\[ \ker \varphi = { g \in G : \varphi(g) = e' } \tag{16} \]
For the sign map above, the kernel is the set of rotations, \({e, r, r^2}\) (which is the normal subgroup we started this post with). This is not a coincidence. In fact, every kernel is normal. Take \(k \in \ker\varphi\) and any \(g \in G\), and just compute:
\[ \varphi(gkg^{-1}) = \varphi(g)\varphi(k)\varphi(g)^{-1} = \varphi(g) e' \varphi(g)^{-1} = e' \tag{17} \]
so \(gkg^{-1} \in \ker\varphi\). Conjugation can't get out of the kernel, so the kernel is normal. The converse holds too: every normal subgroup is the kernel of something, namely the map \(G \to G/N\) sending \(g \mapsto gN\). So "normal subgroup" and "kernel of a homomorphism" describe exactly the same objects.
1.6 The first isomorphism theorem
Which brings us to this interesting result. If \(\varphi: G \to G'\) is a homomorphism, then
\[ G / \ker\varphi \cong \operatorname{im}\varphi \tag{18} \]
The proof is short. Map the coset \(g\ker\varphi\) to \(\varphi(g)\). It's well defined, because replacing \(g\) by \(gk\) with \(k \in \ker\varphi\) gives \(\varphi(gk) = \varphi(g)\varphi(k) = \varphi(g)\), the same answer. It's injective, because \(\varphi(g_1) = \varphi(g_2)\) implies \(\varphi(g_1^{-1}g_2) = e'\), so \(g_1^{-1}g_2 \in \ker\varphi\), so \(g_1\) and \(g_2\) share a coset. And it's onto the image by construction. And we can check this: the sign map has kernel \({e, r, r^2}\) and image \({+1, -1}\), so the theorem says
\[ D_3 / { e, r, r^2 } \cong \mathbb{Z}_2 \tag{19} \]
which is exactly the quotient we computed. Why care? Simply put, it says a homomorphism is completely described by what is in its kernel. Fix the kernel and you've fixed the image up to isomorphism. When we get to representations and start mapping abstract groups into matrices, this tells us what the matrices can look like.