Prerequisites: Quotient group, kernel of a homomorphism, normal subgroup, basic category theory.
Apologize for the handwriting, since it’s hard for quarto to support tikz-cd package.
Takeaways
- You gotta think of a homomorphism whenever there is a quotient.
- Homomorphisms are WAAAAY more important than groups themselves!
- First isomorphism theorem: Universal property of quotients, most important.
- Second isomorphism theorem: How subgroups behave under projections.
- Third isomorphism theorem: Just a diagram commutes.
Normal Subgroup \(\iff\) Kernel
In this section, I will introduce to you a very important mathematical habit: always think of a homomorphism whenever there is a quotient. This helps me a lot in not only understanding the three isomorphism theorems, but also in various topics such as projective geometry, cohomology, etc. First, let’s look at some concepts, if you are familiar with them, you can skip this section.
Center and Centralizer
There are two very similar concepts: center of a group and conjugation operation. Center can be generalized to the concept of a centralizer. Conjugation is related to the concept of a normal subgroup and normalizer.
We know that not every group is abelian, but there are some elements in a group that commute with every other element in the group. We collect them to form a set. In fact, not only a set, but also a subgroup, called the center of the group.
Definition 1 The center \(Z(G)\) of a group \(G\) is: \[
\begin{aligned}
Z(G) :=& \{ z \in G \mid \forall g \in G, zg = gz \} \\
=& \{ z \in G \mid \forall g \in G, z = gzg^{-1} \}.
\end{aligned}
\]
Here are some properties of the center, you can think about them yourself:
- \(Z(G) \trianglelefteq G\). (See Definition 4 for the definition of normal subgroup.)
- \(Z(G) = G\) iff \(G\) is abelian.
- \(Z(G)\) is itself abelian.
If we do not want to make it commutes with every element in the group, but only with some elements in a set \(S\) (not necessarily a subgroup of \(G\)), this generalize to the definition of the centralizer of \(S\) in \(G\):
1 \(S \le G\) means \(S\) is a subgroup of \(G\). \(S \subseteq G\) means \(S\) is a subset of \(G\).
Definition 2 The centralizer \(C_G(S)\) of a set \(S\) \((S \subseteq G)\) of group \(G\) is: \[
\begin{aligned}
C_G(S) :=& \{ c \in G \mid \forall s \in S, cs = sc \} \\
=& \{ c \in G \mid \forall s \in S, c = scs^{-1} \}.
\end{aligned}
\]
Closure:
Let \(a, b \in C_G(S)\), i.e., \(\forall s \in S, as = sa, bs = sb\). We want to show that \(ab \in C_G(S)\): \[
(ab)s = a(bs) = a(sb) = (as)b = (sa)b = s(ab) \implies ab \in C_G(S), \forall s \in S.
\]
Associativity: Inherited from \(G\).
Identity: \(e_G \in C_G(S)\) because \(\forall s \in S, es = se = s\).
Inverse: Let \(a \in C_G(S)\), i.e., \(\forall s \in S, as = sa\). We want to show that \(a^{-1} \in C_G(S)\): \[
\begin{aligned}
as = sa & \implies a^{-1}as = a^{-1}sa \\
& \implies s = a^{-1}sa \\
& \implies sa^{-1} = a^{-1}saa^{-1} \\
& \implies sa^{-1} = a^{-1}s \\
& \implies a^{-1} \in C_G(S).
\end{aligned}
\]
Clearly, the center is a special case of the centralizer: \[
Z(G) = C_G(G).
\]
Conjugation
The sandwich operation \(gzg^{-1}\) in Definition 1 pops out frequenctly in math. We give it a name: conjugation of \(z\) by \(g\).
2 Similar matrices, quarternion representation of 3D rotation, etc.
Definition 3 \(a, b \in G\) conjugate \(:\iff\) \(\exists g \in G\) such that \(a = gbg^{-1}\).
It’s easy to show that conjugation defines an equivalence relation on \(G\):
To show conjugation defined in Definition 3 is an equivalence relation \(\sim\), we need to show it is reflexive, symmetric, and transitive.
Reflexive: \(\exists e \in G, a = eae^{-1}\), hence \(a \sim a\).
Symmetric: \[
\begin{aligned}
a \sim b & \iff \exists g \in G, a = gbg^{-1} \\
& \iff \exists g^{-1} \in G, b = g^{-1}ag \\
& \iff b \sim a.
\end{aligned}
\]
Transitive: \[
\begin{aligned}
a \sim b, b \sim c & \iff \exists g, h \in G, a = gbg^{-1}, b = hch^{-1} \\
& \iff \exists g, h \in G, a = ghch^{-1}g^{-1} \\
& \iff \exists gh \in G, a = (gh)c(gh)^{-1} \\
& \iff a \sim c.
\end{aligned}
\]
A Natural Homomorphism
Proposition 1 Let \(G\) be a group and \(f \in \operatorname{Hom}(G, \operatorname{Aut}(G))\), where \(f(g)\) is defined by conjugation by \(g\). Then \[
\ker f = Z(G).
\]
Conjugation by \(g\) (denoted \(\operatorname{conj}_g\)) is in \(\operatorname{Aut}(G)\).
We need to check that \(\operatorname{conj}_g\) gives an isomorphism from \(G\) to itself.
Claim: \(\forall g \in G, f(g) \in \operatorname{End}(G)\).
\[
f(g)(h h') = g(h h')g^{-1} = ghg^{-1}gh'g^{-1} = f(g)(h) f(g)(h').
\]
Claim: \(f(g)\) is bijective.
\[
f^{-1}(g) = f(g^{-1}).
\]
\(f\) is indeed a homomorphism.
Take \(\forall g, g' \in G\), then: \[
\begin{aligned}
f(g g')(h) &= gg'h(gg')^{-1} \\
&= gg'hg'^{-1}g^{-1} \\
&= f(g) \circ f(g')(h).
\end{aligned}
\]
\(\ker f = Z(G)\).
\[
\begin{aligned}
\ker f &= \{g \in G : \forall h \in G, ghg^{-1} = h\} \\
&= \{g \in G : gh = hg \text{ for all } h \in G\} \\
&= Z(G).
\end{aligned}
\]
Proposition 2 Under the condition in Proposition 1, \[
\operatorname{im} f \triangleleft \operatorname{Aut}(G).
\]
Define the inner automorphism group to be: \[
\operatorname{Inn}(G) := \operatorname{im} f,
\] and naturally the Outer automorphism group to be: \[
\operatorname{Out}(G) := \frac{\operatorname{Aut}(G)}{\operatorname{Inn}(G)}.
\] Then we have the following exact sequence: \[
1 \xrightarrow[]{} Z(G) \hookrightarrow G \xrightarrow[]{f} \operatorname{Aut}(G) \twoheadrightarrow \operatorname{Out}(G) \xrightarrow[]{} 1.
\]
We omitted the proof, but visually,
Normal Subgroup and Normalizer
In fact, we could weaken the condition of \(z = gzg^{-1}\) in Definition 1 and \(c = scs^{-1}\) in Definition 2 to \(N = gNg^{-1}\) and \(S = gSg^{-1}\) to get the definition of a normal subgroup \(N\) and a normalizer \(N_G(S)\) respectively. This is different, because we no longer need pointwise commutativity but only commutativity as a set. In other words, the element after conjugation does not need to be exactly itself necessarily as long as it remains in some set.
Definition 4 A subgroup \(N\) of a group \(G\) is normal (denoted \(N \trianglelefteq G\)), iff \(\forall g \in G, N = gNg^{-1}\), i.e., \[
\forall g \in G, \exists n \in N, \forall m \in N, gmg^{-1} = n,
\] i.e., normal subgroups are invariant under any conjugation.
Definition 5 The normalizer \(N_G(S)\) of a set \(S\) \((S \subseteq G)\) of group \(G\) is: \[
\begin{aligned}
N_G(S) :=& \{ g \in G \mid gS = Sg \} \\
=& \{ g \in G \mid S = gSg^{-1} \}.
\end{aligned}
\]
Their relationship is: If \(S \trianglelefteq G\), then \(N_G(S) = G\).
The following two results are important:
Kernels are normal
Lemma 1 Let \(G, H\) be groups and \(\varphi \in \operatorname{Hom}(G, H)\). Then \(\ker(\varphi) \trianglelefteq G\), \(\operatorname{im}(\varphi) \le H\).
We just need to show that \(\forall k \in \ker(\varphi)\), the conjugate of \(k (gkg^{-1})\) is also in \(\ker(\varphi)\), i.e., \[
\varphi(gkg^{-1}) = e.
\] This is trivial.
Normal subgroups are kernels
Lemma 2 Let \(N \trianglelefteq G\). Then \(\exists \varphi \in \operatorname{Hom}(G, H)\) s.t. \(N = \ker(\varphi)\).
HINT: Take \(\varphi: G \to G/N\), the natural projection from \(G\) to the group of cosets (called the quotient group \(G/N\)).
As shown in Figure 2, any group homomorphism \(\varphi: G \to H\) factors uniquely through the quotient group \(G/N\), where \(N = \ker(\varphi)\). The unique embedding \(\bar{\varphi}\) implies the universal property of the quotient group (the pair \((G/N, \pi)\) could be viewed as a initial object of a coslice category).
3 A coslice category is the dual of a slice category, which is a special case of a comma category.
So we have establish the fact that quotients and group homomorphisms are equivalent!
Isomorphism Theorem I
Figure 2 could be viewed as part of the canonical decomposition of a group homomorphism:
The red box in Figure 3 is exactly the first isomorphism theorem:
Theorem 1 Let \(G\) and \(H\) be groups and let \(\varphi: G \to H\) be a homomorphism. Then: \[
G/\ker(\varphi) \simeq \operatorname{im}(\varphi)
\]
The rest of the isomorphism theorems largely relies on this first theorem.
Isomorphism Theorem II
Lemma 3 Let \(G\) be a group and \(N \trianglelefteq G\), \(S \le G\). Then: \[
N \trianglelefteq SN \le G \ge S \trianglerighteq S \cap N.
\]
\(SN \le G\):
- Nonempty: Trivial.
- Closure: Let \(s_1n_1, s_2n_2 \in SN\), we want to show that \((s_1n_1)(s_2n_2) \in SN\): \[
(s_1n_1)(s_2n_2) = s_1(n_1s_2)n_2.
\] How to deal with the middle term? Since \(N \trianglelefteq G\), we have \(n_1s_2 = sn_1\) for some \(s \in N\). Then: \[
s_1(n_1s_2)n_2 = s_1(sn_1)n_2 = (s_1s)(n_1n_2) \in SN.
\]
- Inverses: Let \(sn \in SN\), we want to show that \((sn)^{-1} = n^{-1}s^{-1} \in SN\). Again use the fact that \(N \trianglelefteq G\): \[
\exists s' \in N, n^{-1}s^{-1} = s'n^{-1} \in SN.
\]
The rest can be proved by this diagram:
Theorem 2 Let \(G\) be a group and \(N \trianglelefteq G\), \(S \le G\). Then: \[
\frac{SN}{N} \simeq \frac{S}{S \cap N}
\]
As shown in Figure 4, we have:
- \(N \trianglelefteq SN\), since \(N = \ker(\varphi|_{SN})\). By the first isomorphism theorem, we have \(\frac{SN}{N} \simeq \varphi (SN)\).
- \(S \cap N \trianglelefteq S\), since \(S \cap N = \ker(\varphi|_{S})\). By the first isomorphism theorem, we have \(\frac{S}{S \cap N} \simeq \varphi (S)\).
Plus, \(\varphi (SN) = \varphi(S)\), therefore: \[
\frac{SN}{N} \simeq \varphi (SN) = \varphi(S) \simeq \frac{S}{S \cap N},
\] i.e., \[
\frac{SN}{N} \simeq \frac{S}{S \cap N}.
\]
Isomorphism Theorem III
Theorem 3 Let \(G\) be a group and \(N \trianglelefteq G\), \(K \trianglelefteq G\), \(N \subset K\). Then \(N \trianglelefteq K\) and \[
\frac{G/N}{K/N} \simeq \frac{G}{K}
\]
To prove this theorem, we need to prove a lemma:
Lemma 4 Let \(G\) be a group, \(K \trianglelefteq G\), \(\varphi \in \operatorname{Hom}(G, H)\). The image of \(K\) under \(\varphi\) is normal in \(H\), i.e., \[
\varphi(K) \trianglelefteq \operatorname{im} \varphi
\]
We need to show \(\forall h \in \operatorname{im} \varphi\) and \(\forall k \in K\), \(h\varphi(k)h^{-1} \in \varphi(K)\).
Let:
- \(h \in \operatorname{im}(\varphi)\), so there exists \(g \in G\) such that \(h = \varphi(g)\),
- \(k \in K\), so \(\varphi(k) \in \varphi(K)\)
Now compute: \[
h \varphi(k) h^{-1} = \varphi(g) \varphi(k) \varphi(g)^{-1}
= \varphi(g k g^{-1}) \quad \text{(since \( \varphi \) is a homomorphism)}
\]
But since $ K G $, we know \(g k g^{-1} \in K\), so: \[
\varphi(g k g^{-1}) \in \varphi(K).
\] Therefore: \[
h \varphi(k) h^{-1} \in \varphi(K).
\]
So \(\varphi(K)\) is closed under conjugation by elements of \(\operatorname{im}(\varphi)\), i.e.,
\[
\varphi(K) \trianglelefteq \operatorname{im}(\varphi).
\]
Now the proof of the third isomorphism theorem can be proved by this diagram:
Figure 5 can be visually represented by this cartoon:
by the first isomorphism theorem, we have: \[
G/N \simeq \sigma_1 (G) \triangleright \sigma_1 (K) \simeq K/N.
\] Therefore, \[
\frac{G/N}{K/N} = \frac{\sigma_1 (G)}{\sigma_1 (K)} \simeq \sigma_2 \circ \sigma_1 (G) =: \sigma (G) \simeq \frac{G}{K}.
\]
