Basic definitions of a Topological Space

For an explanation of this type of post, go here.

In this post I’m basically just following along with General Topology by Kelley [1] aided by Wikipedia. Absolutely no thought went into this reference choice so I hope it isn’t impossibly difficult. I will also break my rule about posting solutions to exercises here because I don’t have confidence in making up my own exercises and I think this book is sufficiently old (published 1955) that no one is using it in class anymore. Elementary point-set topology, like elementary analysis, is at least fun to think about before it gets too abstract for my abilities. As always I use my reference for the statement of the theorems and the supplementary text and try to prove the theorems myself.

With topology I start from the very beginning, so I hope all readers with be able to follow along to some degree. The first question I had to ask is what the basic unit of study is. It turns out it’s a topological space. How is it defined? Apparently in a bunch of different ways, as we are about to see.

We first define a topological space in the following manner:

Definition 1 (Topology by Open Sets) A topological space ${(X,\mathcal{T})}$ is an ordered pair of a set and a family of subsets with the following properties:

1. Both ${X}$ and ${\emptyset}$ are in ${\mathcal{T}}$.
2. For any ${U,V \in \mathcal{T}}$, ${U \cap V \in \mathcal{T}}$.
3. The union of a family of subsets of ${\mathcal{T}}$ is also in ${\mathcal{T}}$.

The set ${X}$ is the space (universe) and ${\mathcal{T}}$ is called the topology on ${X}$.

The second stipulation of course implies that the intersection of any finite number of sets in ${\mathcal{T}}$ is still in ${\mathcal{T}}$. By contrast the third stipulation must not only hold for finite sets but also countably and uncountably infinite sets as well. Interestingly enough there are a bunch of ways to get to an equivalent definition according to Wikipedia, but we’ll stick to this one to begin with. This is the definition everyone seems to use. Kelley actually works in a marginally different way – he doesn’t have the first condition at all and just defines ${X}$ to be the union of all sets in ${\mathcal{T}}$. Therefore ${X}$ is in ${\mathcal{T}}$ by definition. I find this a little strange because he talks about pairs ${(X,\mathcal{T})}$ just like everyone else, but in the case of his definition ${X}$ is already completely determined by ${\mathcal{T}}$, so I don’t really understand why this is done. His third condition is also worded slightly differently: he says that “the union of any subfamily of sets of ${\mathcal{T}}$ is also in ${\mathcal{T}}$.” Under this phrasing he goes on to say that this implies that ${\emptyset \in \mathcal{T}}$ since ${\{ \}}$ is a subfamily of the family of sets and its union is ${\emptyset}$. I don’t want to get too technical for no reason so I’m just going to stick with the majority here. It’s just a matter of what to verify when showing something is a topology anyway.

The sets in ${\mathcal{T}}$ are called the open sets. Technically openness is a property of the topology, so they are ${\mathcal{T}}$-open, but I think it will be mostly clear from context. If there are two topologies in question then we’ll specify which we are talking about. A go-to example of a topological space is of course the real numbers. The usual topology for the real numbers refers to the family of sets which contain an open interval about each of their points. First we’ll better define an open interval and verify that it in fact forms a topology:

Proposition 2 Let an open interval ${I}$ be a subset of ${\mathbb{R}}$ such that for every ${x \in I}$ there exists ${a,b \in \mathbb{R}}$ such that ${(a,b)}$ contains ${x}$ and is contained in ${I}$. Then the family of sets defined by all open intervals on ${\mathbb{R}}$ is a topology.

Proof: Since there does not exist an ${x \in \emptyset}$ it is included in our family of sets. Similarly if we consider ${\mathbb{R}}$ then any ${a,b}$ such that ${a < x < b}$ is contained in ${\mathbb{R}}$ so this too is in our family of sets.

If ${I_1}$ and ${I_2}$ are disjoint open intervals then their intersection forms the empty set, so that case is fine. If ${I_1}$ and ${I_2}$ are open intervals where there exists an ${x \in I_1 \cap I_2}$, let ${x \in (a_1,b_1) \subset I_1}$ and ${x \in (a_2,b_2) \subset I_2}$. Then ${\max(a_1,a_2) < x < \min(b_1,b_2)}$ is contained in ${I_1 \cap I_2}$, so this is again open.

Let ${U = \bigcup_{a \in A} U_a}$ be the union of some family of open sets where ${A}$ is an index set. Then for any ${x \in U}$ we know ${x \in U_a}$ for some ${a \in A}$, and so ${U_a}$ must contain an interval ${(a,b)}$ that contains ${x}$ which in turn must be contained in ${U}$, so ${U}$ is again open. $\square$

That proof is so easy Kelley didn’t even bother to include it, but I figured I’d write it up to decrease the probability I was misunderstanding everything. Of course under this definition open intervals conveniently correspond to open sets, and since a set is closed iff it is the complement of some open set, closed intervals correspond to closed sets. To my dismay (and the cost of about an hour of staring at a piece of paper), this particular naming scheme does not hold up in more general cases. The usual topology does not allow sets to be both open and closed, excepting the empty set and the entire space, but a generic topology allows this for more than just the two sets.

neighborhood of a point ${x}$ is a set ${U}$ which contains an open set containing ${x}$. A neighborhood system of a point ${x}$ is the family of all neighborhoods of ${x}$. This sets up Kelley’s first two theorems:

Theorem 3 A set is ${U}$ open if and only if it contains a neighborhood of ${x}$ for every ${x \in U}$.

Proof: If a set ${U}$ is open then it itself is a neighborhood of any ${x \in U}$, so that direction is simple. If ${U}$ contains a neighborhood of ${x}$ for every ${x \in U}$, it also contains an open set which contains ${x}$. Consider the union of all these open sets, call it ${V}$, which is by definition again open. Then clearly ${U \subset V}$ since ${x \in U}$ implies ${x \in V}$. We also have ${V \subset U}$ since ${V}$ is just a union of subsets of ${U}$, so ${U=V}$ and ${U}$ is open. $\square$

Theorem 4 If ${\mathcal{U}}$ is the neighborhood system of a point ${x}$, then finite intersections of members of ${\mathcal{U}}$ belong to ${\mathcal{U}}$, and each set which contains a member of ${\mathcal{U}}$ belongs to ${\mathcal{U}}$.

Proof: The latter statement is kind of obvious, because a set which contains a neighborhood of ${x}$ must contain an open set containing ${x}$, so it is also a neighborhood of ${x}$.

Let ${U,V}$ be neighborhoods of ${x}$. Then there exist open sets ${U',V'}$ containing ${x}$ such that ${U' \subset U}$ and ${V' \subset V}$. We see that ${U' \cap V'}$ is open by definition and ${U' \cap V' \subset U \cap V}$, so ${U \cap V}$ contains an open set containing ${x}$ and is therefore a neighborhood of ${x}$ $\square$

This brings us to another definition of a topological space, which we then show to be equivalent to our first definition.

Definition 5 (Topology by Neighborhoods) Let ${\mathcal{U}}$ be a function which assigns to each ${x \in X}$ a non-empty family of sets ${\mathcal{U}_x}$ such that the following properties hold:

1. If ${U \in \mathcal{U}_x}$, then ${x \in U}$.
2. If ${U}$ and ${V}$ are members of ${\mathcal{U}_x}$, then ${U \cap V \in \mathcal{U}_x}$.
3. If ${U \in \mathcal{U}_x}$ and ${U \subset V}$, then ${V \in \mathcal{U}_x}$.
4. If ${U \in \mathcal{U}_x}$, then there is a member ${V}$ of ${\mathcal{U}_x}$ such that ${V \subset U}$ and ${V \in \mathcal{U}_y}$ for each ${y \in V}$.

Then the family ${\mathcal{T}}$ of all sets ${U}$ such that ${U \in \mathcal{U}_x}$ whenever ${x \in U}$ is a topology on ${X}$.

Exercise 1 (1B(a) from Kelley) Show that if ${(X,\mathcal{T})}$ is a topological space and for each ${x}$ let ${U_x}$ be the family of all neighborhoods of ${x}$. Then the four conditions in the above definition hold.

Proof: We’ll go down the list

1. If ${U \in \mathcal{U}}$ then it’s a neighborhood of ${x}$ so by definition it contains an open set which contains ${x}$. Therefore it contains ${x}$.
2. This is our second theorem.
3. If ${U \in \mathcal{U}_x}$, then ${U}$ contains an open set containing ${x}$. If ${U \subset V}$ then ${V}$ contains that same open set containing ${x}$, so ${V \in \mathcal{U}_x}$.
4. We can choose ${V}$ to be the open set contained in ${U}$ which contains ${x}$. Clearly ${V \subset U}$ and ${V \in \mathcal{U}_x}$. Since ${V}$ is itself an open set it must be a neighborhood for all its points. $\square$

So these four conditions are implied by our open sets definition. As you might expect, the second part of the exercise is to show that these four conditions imply our open sets definition.

Exercise 2 (1B(b) from Kelley) If ${\mathcal{U}}$ is a function which assigns to each ${x \in X}$ a non-empty family of sets ${\mathcal{U}_x}$ satisfying the first three conditions in the previous exercise, then the family ${\mathcal{T}}$ of all sets ${U}$, such that ${U \in \mathcal{U}_x}$ whenever ${x \in U}$, is a topology on ${X}$. If the fourth condition is satisfied, then ${\mathcal{U}_x}$ is precisely the neighborhood system of ${x}$ relative to the topology ${\mathcal{T}}$.

Proof: A more explicit definition of ${\mathcal{T}}$ is ${\{U : x \in U \Rightarrow U \in \mathcal{U}_x\}}$.

The empty set is trivially in ${\mathcal{T}}$ and the entire space is as well by the third condition, since it contains all sets (and ${U_x}$ is non-empty). Let ${U,V}$ be sets in ${\mathcal{T}}$. Let ${x}$ be a point in ${U \cap V}$, so ${x \in U}$ and ${x \in V}$. Since ${U,V \in \mathcal{T}}$ we have ${U,V \in \mathcal{U}_x}$. Then by the second condition we have ${U \cap V \in \mathcal{U}_x}$ so ${x \in U \cap V \Rightarrow U \cap V \in \mathcal{U}_x}$ and ${U \cap V \in \mathcal{T}}$. Now let ${K = \bigcup_{a \in A}{W_a}}$ with ${W_a \in \mathcal{T}}$ for all ${a \in A}$ (an index set). We know ${x \in K}$ means that ${x \in W_a}$ for some ${a}$, so ${W_a \in \mathcal{U}_x}$ since ${W_a \in \mathcal{T}}$. But ${W_a}$ is a subset of ${K}$ so by the third condition ${K \in \mathcal{U}_x}$.

Finally, if we also have the fourth condition, we can show that ${\mathcal{T} = \{U : x \in U \Rightarrow U \in \mathcal{U}_x\}}$ is the neighborhood system of ${x}$.

If ${T \in \mathcal{T}}$ then if ${x \in T}$ we have ${T \in \mathcal{U}_x}$. By the fourth condition there exists a ${V}$ which is open (by theorem 3) and ${V \subset T}$, so ${T}$ is a neighborhood of ${x}$ (and is in the neighborhood system). If we have a set ${V}$ which is a neighborhood of ${x}$ then it contains an open set ${Y}$ which contains ${x}$. We know that ${Y \in \mathcal{U}_x}$ by the fourth condition so if ${x \in V}$ then ${V \in \mathcal{U}_x}$ by the third condition. Thus ${x \in V}$ implies ${V \in \mathcal{U}}$ so ${V \in \mathcal{T}}$. We conclude that ${\mathcal{T}}$ is the neigbhorhood system of ${x}$. $\square$

We’re going to run through a few definitions and theorems now to prevent this from getting too long. An accumulation point ${x}$ of a subset ${A \subset \mathcal{T}}$ is a point for which all neighborhoods of ${x}$ intersect ${A}$ at a point other than ${x}$. This is just a step toward the way the neighborhoods definition of a topology expresses closed sets. A subset of a topological space is closed if and only if it contains all its accumulation points. These are of course all slightly generalized terms of what people learn in the first few weeks in analysis. The closure ${\bar{A}}$ of a set ${A}$ is the intersection of all the closed sets containing ${A}$. This leads to a theorem which I will write up the proof for, having skipped enough of them already. Unfortunately I got tripped up on this one so I had to go use Kelley’s proof, which starts with this preliminary theorem:

Theorem 6 The union of a set and the set of its accumulation points is closed.

Proof: For some ${x \notin A \cup acc(A)}$ let ${U}$ be an open set containing ${x}$ and not intersecting ${A}$, which exists because ${x \notin acc(A)}$. In fact, no points of ${U}$ are in ${A}$ by definition and no points are in ${acc(A)}$ since ${U}$ is a neighborhood of all its points. Taking the union of all such ${U}$ over all ${x \notin A \cup acc(A)}$ we get an open set which is the complement of ${A \cup acc(A)}$, so ${A \cup acc(A)}$ must be closed. $\square$

This makes the second half of the next theorem, the part I got stuck on, obvious.

Theorem 7 The closure of any set is the union of the set and the set of its accumulation points

Proof: Let ${acc(A)}$ denote the set of accumulation points of ${A}$. If ${x \in A}$ then it’s obviously in the closure of ${A}$ since closed (or any) sets containing ${A}$ must contain ${x}$. Similarly if ${x \in acc(A)}$ then ${x}$ must be an accumulation point of a set containing ${A}$, and since closed sets must contain all their accumulation points it must be in any closed set containing ${A}$. So if ${x \in A \cup acc(A)}$ then ${x \in \bar{A}}$.

If ${x \in \bar{A}}$ then it’s in every closed set containing ${A}$. Since ${A \cup acc(A)}$ is closed by the previous theorem and obviously contains ${A}$, ${x \in A \cup acc(A)}$. $\square$

This gives us yet another way to define a topological space. A closure operator on ${X}$ is an operator which assigns ${A}$ to ${A^c}$ (not to be confused with complement) such that the following axioms are satisfied:

Definition 8 (Kuratowski Closure Axioms) Let ${c}$ be a closure operator on ${X}$.

1. We have ${\emptyset^c = \emptyset}$.
2. For all ${A}$, ${A \subset A^c}$.
3. For all ${A}$, ${(A^c)^c = A^c}$.
4. For all ${A,B}$ we have ${(A \cup B)^c = A^c \cup B^c}$.

This gives us our definition:

Theorem 9 (Topology by Closure Operator) Let ${c}$ be a closure operator on ${X}$, let ${\mathcal{F}}$ be the family of all subsets ${A}$ of ${X}$ such that ${A^c = A}$, and let ${\mathcal{T}}$ be the family of complements of members of ${\mathcal{F}}$. Then ${\mathcal{T}}$ is a topology on ${X}$ and ${A^c = \bar{A}}$ (with respect to ${\mathcal{T}}$) for all ${A \ subset X}$.

Proof: We know that ${\emptyset \in \mathcal{F}}$ by the first definition, so ${X \in \mathcal{T}}$. Similarly since ${X^c = X}$ since ${X \subset X^c}$. Since we’re dealing with closed sets directly it’s probably easier to apply demorgan’s to the open sets definition of a Topology. We then need to show that the union of two sets in ${\mathcal{F}}$ is in ${\mathcal{F}}$ and the intersection of an arbitrary number of sets in ${\mathcal{F}}$ is in ${\mathcal{F}}$. For ${F_1, F_2 \in \mathcal{F}}$, we have ${(F_1 \cup F_2)^c = F_1^c \cup F_2^c = F_1 \cup F_2}$ using the fourth axiom, so ${F_1 \cup F_2 \in \mathcal{F}}$. If ${K = \bigcap_{a \in A}{F_a}}$ where ${A}$ is an index set and ${F_a \in \mathcal{F}}$ for all ${a}$, we have the following chain: ${K^c = (\bigcap{F_a})^c \subset \bigcap(F_a^c) = \bigcap(F_a) = K}$, so ${K^c \subset K}$. By the second axiom ${K \subset K^c}$, so ${K = K^c}$ and ${K \in \mathcal{F}}$.

Finally we want to show that ${A^c}$ is actually just ${\bar{A}}$. I took the massive implied hint here and used the third axiom since it hasn’t come up yet. By that axiom we see that ${A^c}$ is actually in ${\mathcal{F}}$. But ${\bar{A}}$ is the intersection of the closed sets containing ${A}$, i.e. the sets in ${\mathcal{F}}$ containing ${A}$, and since ${A \subset A^c}$ we see that ${\bar{A} \subset A^c}$. For the other direction we note that ${\bar{A} = A \cup B}$ for some (possibly empty) ${B}$. So ${\bar{A}^c = (A \cup B)^c = A^c \cup B^c \supset A^c}$, and since ${\bar{A}}$ is actually in ${\mathcal{F}}$ because it’s an intersection of elements of ${\mathcal{F}}$, we have ${\bar{A}^c = \bar{A}}$. Putting the two together yields ${A^c \subset \bar{A}}$, completing the proof. $\square$

To conclude we’ll look at one more way to specify a topological space after the relevant definitions. The interior of a set ${A}$ is the set of ${x \in A}$ such that ${A}$ is a neighborhood of ${x}$. It’s denoted  $A^i$. I tend to think of the interior operator as the opposite of the closure operator, and indeed it can be shown that the interior is the largest open subset of ${A}$, as opposed to the smallest closed set containing ${A}$. This kind of duality holds up pretty well. The interior operator, as you might imagine, takes ${A}$ to ${A^i}$. Now we’re going to define a topology one last time, based on this exercise:

Exercise 3 (1C in Kelley) If ${i}$ is an operator which carries subsets of ${X}$ into subsets of ${X}$, and ${\mathcal{T}}$ is the family of all subsets such that ${A^i = A}$, under what conditions will ${\mathcal{T}}$ be a topology for ${X}$ and ${i}$ the interior operator relative to this topology?

Definition 10 (Topology by Interior Operator) Let \textsuperscript{${i}$} be an interior operator on ${X}$ where ${\mathcal{T}}$ is the family of subsets such that ${A^i = A}$, and let the following conditions hold:

1. We have ${X^i = X}$.
2. For all ${A}$, ${A \supset A^i}$.
3. For all ${A}$, ${(A^i)^i = A^i}$.
4. For all ${A,B}$, ${(A \cap B)^i = A^i \cap B^i}$.

Then ${\mathcal{T}}$ is a topology on ${X}$ and ${i}$ is its interior operator.

It did not take a lot of creativity to come up with those conditions, of course. Since this post has more than its fair share of uninteresting proofs and this one is extremely similar to the previous one, I will omit it.

That admittedly got really long and dry. I wanted to do all the really mundane and boring exercises for practice and confirmation of understanding. I wouldn’t even recommend a reader go too carefully through the proofs to learn (if you are then I recommend reading a real resource). I am, however, hopeful that in coming posts we’ll get to some more tangible results. In those perhaps there will be more value in cursory reading.

Source: [1] Kelley, John. General Topology. Published 1955 by D. Van Nostrand Company, Inc.
Used for the statement of definitions, theorems, comments, and theorem proofs where specified.

The Isomorphism Theorems

For a short explanation of this type of post, go here.

In this post I’m using Dummit and Foote [1] and Wikipedia for reference. As a minor disclaimer some of Dummit and Foote’s proofs are more general. I made a few extra assumptions that at least in my understanding still give us results that are sufficiently powerful for our purposes. As with all these theorems the proofs are simple. I have omitted large chunks, mostly the parts I consider verification, but if you’re following along I would recommend at least mentally checking off properties. Dummit and Foote actually introduce quotient groups using homomorphisms rather than defining normal subgroups and then quotient groups but while I think that’s a reasonable approach I don’t think it’s how most people first learned the concept. I went with the traditional approach of beginning with normal subgroups.

Anyway, let’s get on to the material. First a brief and common lemma will make my life a bit easier:

Lemma 1 When ${G}$ is a group and ${K \leq G}$, we have ${g_1K = g_2K \Longleftrightarrow g_2^{-1}g_1 \in K}$

Proof: If ${g_2^{-1}g_1 \in K}$ then ${g_2^{-1}g_1 = k}$ for some ${k \in K}$, and we have ${g_1 = g_2k}$, so ${g_1K = g_2kK = g_2K}$. Conversely if ${g_1K = g_2K}$ then for some ${k_1,k_2 \in K}$ we have ${g_1k_1 = g_2k_2}$ so ${g_1 = g_2k_2k_1^{-1} \Longrightarrow g_2^{-1}g_1 = k_2k_1^{-1} \in K}$. $\square$

As a last comment, in case this notation isn’t standard, Dummit and Foote use ${H \leq G}$ to say ${H}$ is a subgroup of ${G}$, ${N \trianglelefteq G}$ to say ${N}$ is a normal subgroup of ${G}$, and ${K \cong G}$ to say ${K}$ is isomorphic to ${G}$.

Theorem 2 (The First Isomorphism Theorem) Let ${\varphi: G \rightarrow H}$ be a surjective homomorphism. Then ${\ker(\varphi) \trianglelefteq G}$ and ${G / \ker(\varphi) \cong H}$. For ease of writing we’ll let ${K}$ be ${\ker(\varphi)}$.

Proof: First we’ll verify that ${\ker(\varphi) \trianglelefteq G}$.

• Closure: For ${k_1,k_2 \in K}$ we have ${\varphi(k_1k_2) = \varphi(k_1)\varphi(k_2) = e_He_H = e_H}$, so ${k_1k_2 \in K}$.
• Identity: Since for any ${g \in G}$ we have ${\varphi(g) = \varphi(e_Gg) = \varphi(e_G)\varphi(g)}$, ${\varphi(e_G) = e_H}$ and so ${e_H \in K}$.
• Inverses: For any ${k \in K}$ we have ${\varphi(k)\varphi(k^{-1}) = \varphi(kk^{-1}) = \varphi(e_G) = e_H}$, but since ${\varphi(k) = e_H}$ we also have that ${\varphi(k^{-1}) = e_H}$ and so ${k^{-1} \in K}$.

To show that ${G/K \cong H}$ we just need to show that ${\phi: G/K \rightarrow H}$ where ${\phi(gK) = \varphi(g)}$ is well-defined, a homomorphism, injective, and surjective.

• Well-defined: If ${g_1K = g_2K}$, then ${g_2^{-1}g_1 \in K}$ by lemma 1 and so ${\varphi(g_2^{-1}g_1) = e_H}$, ${\varphi(g_2^{-1})\varphi(g_1) = e_H}$, ${\varphi(g_2^{-1})\varphi(g_1) = e_H}$, and finally ${(\varphi(g_2)^{-1})^{-1} = \varphi(g_1)}$, so ${\varphi(g_2) = \varphi(g_1)}$ which means that ${\phi}$ is well defined.
• Homomorphism: Since ${\phi(g_1K)\phi(g_2K) = \varphi(g_1)\varphi(g_2) = \varphi(g_1g_2) = \phi(g_1g_2K) = \phi((g_1K)(g_2K))}$, ${\phi}$ is a homomorphism of groups.
• Injective: This amounts to reversing the proof that ${\phi}$ is well defined: We have ${\phi(g_1K) = \phi(g_2K) \Longrightarrow \varphi(g_1) = \varphi(g_2) \Longrightarrow \varphi(g_1)\varphi(g_2^{-1}) = e_H \Longrightarrow \varphi(g_1g_2^{-1}) = e_H \Longrightarrow g_1g_2^{-1} \in K}$, so we can conclude ${g_1K = g_2K}$ by lemma 1 and so ${\phi}$ is injective.
• Surjective: For any ${h \in H}$ there exists a ${g \in G}$ such that ${\varphi(g) = h}$ since ${\varphi}$ is surjective, and so ${\phi(gK) = \varphi(g) = h}$, so ${\phi}$ is surjective. $\square$

Dummit and Foote also point out an immediate consequence of this theorem that sort of sheds some light on its utility:

Corollary 3 If ${\varphi: G \rightarrow H}$ is a group homomorphism then ${\varphi}$ is injective iff ${\ker(\varphi) = e_G}$.

Proof: If ${\varphi}$ isn’t surjective then we need only consider its image, so we can just assume that it is. If ${\varphi}$ is injective then it’s a bijection so the statement is obvious. If ${\ker(\varphi) = e_G}$ then ${G / \ker(\varphi) = G}$ and so ${\varphi}$ is an isomorphism, so it’s injective. $\square$

A picture I like to maintain in my head looks like this:

Where ${K}$ is ${\ker(\varphi)}$, ${\varphi: G \rightarrow H}$, ${\phi: G/K \rightarrow H}$ is defined as ${\phi(gK) = \varphi(g)}$, and ${\pi: G \rightarrow G/K}$ is defined as ${\pi(g) = gK}$. We can think of ${\varphi}$ as ${\pi \circ \phi}$, and while this doesn’t immediately appear helpful, in the case where ${\ker(\varphi) = 1}$ we can see that this is actually a lot of information about ${\varphi}$, since it fixes ${G/K}$, in this case ${G}$. By relating ${\varphi}$ to this composition we preclude the idea that it might be mapping elements in an unequal fashion because of Lagrange’s theorem. In my mind I think of this as saying that the set of homomorphisms on ${G}$ are in some sense equivalent to the set of normal subgroups, although this is mostly speculation.

The next two theorems heavily utilize the first, so hopefully the result makes sense. We’ll see some application

Theorem 4 (The Second Isomorphism Theorem) Let ${G}$ be a group and let ${S,N \leq G}$ with ${N}$ normal. Then ${SN \leq G}$, ${S \cap N \trianglelefteq S}$, and ${(SN)/N \cong S/(S \cap N)}$.

Proof: The first two claims are a matter of verification, so I’ll skip those. As a result of those claims those we know that ${S \cap N}$ is a normal subgroup of ${S}$ so ${S/(S \cap N)}$ is well defined. For the other quotient group ${N}$ is normal in ${G}$ by assumption and since ${SN \leq G}$ it’s also normal in ${SN}$ (note that this is not guaranteed for some ${F}$ where ${G \leq F}$).

Here’s an outline of what we’re going to do. We want to show that ${S/(S \cap N) \cong (SN)/N}$, and we know that ${G/\ker(\varphi) \cong \varphi(G)}$. So it would be ideal if we could define a ${\varphi: S \rightarrow (SN)/N}$ such that ${\ker(\varphi) = S \cap N}$. If we do this then by the first isomorphism we’ll be done.

An obvious guess at how to define ${\varphi}$ is to let ${\varphi(s) = sN}$. Since all our quotient groups are legitimate this is well-defined and everything, and it’s easy to show that it’s a surjective homomorphism. Really all that remains to be shown is that ${\ker(\varphi) = S \cap N}$. The identity of ${(SN)/N}$ is ${N}$, so the kernel of ${\varphi}$ is the set of elements ${x \in S}$ where ${xN = N}$. By lemma 1 this implies that ${x \in N}$, and so the kernel of ${\varphi}$ is the set of elements in ${N}$ but also in ${S}$ by definition, so ${S \cap N}$. So we have a ${\varphi}$ such that ${\varphi: S \rightarrow (SN)/N}$ with ${\ker(\varphi) = S \cap N}$, so by the first isomorphism theorem we’re done. $\square$

I was actively trying to apply the first isomorphism theorem in that proof. I’m sure there are more direct ways to prove it, but I don’t think this one is too bad. It basically illustrates a situation where if you define a homomorphism you can make some non-obvious statements about the structure of ${G}$. In this case the second isomorphism theorem was here to tell me how to fit the pieces together to make the result seem nice, but the idea is more general. Here’s the relevant part of the lattice diagram of ${G}$, where quotient groups formed by subgroups joined by the red lines are isomorphic.

Dummit and Foote also call this the diamond isomorphism theorem. The third isomorphism theorem, stated next, is another example of showing isomorphism by defining the appropriate homomorphism.

Theorem 5 (The Third Isomorphism Theorem) Let ${G}$ be a group with ${H,K \trianglelefteq G}$ and ${H \leq K}$. Then ${(K/H) \trianglelefteq (G/H)}$ and ${(G/H)/(K/H) \cong G/K}$.

Proof: Again I’ll leave it to the reader to verify the first part. The last part is more fitting pieces into the first isomorphism theorem. Let ${\varphi: G/H \rightarrow K/H}$ be defined by ${\varphi(gH) = kH}$ which can be verified as a surjective homomorphism. Then ${\ker(\varphi)}$ is the set of cosets of the form ${gH}$ such that ${gK \in K}$, by again by lemma 1 this implies that ${g \in K}$, so the kernel is the set of cosets of the form ${kH}$, better known as the elements of ${K / H}$, and so by the first isomorphism theorem we’re done. $\square$

Putting aside the obvious jokes about canceling the ${H}$, after the first isomorphism theorem we pretty much went back to our symbol-manipulating ways. We could work through some examples of groups but I think there is only limited utility in that. A real application of the theorems is more desirable, and these do come up in solvable groups if I recall correctly. This gives me an excuse to get sidetracked in the Galois theory proof that quintics are not solvable in radicals. Personally this is one of my favorite results in all of math, which is why I was looking for an excuse to go learn about it.

Some authors, Dummit and Foote included, have a fourth isomorphism theorem that for a group ${G}$ and normal subgroup ${N}$ shows a bijection between the subgroups of ${G}$ which contain ${N}$ and the subgroups ${G/N}$. I don’t think this is any less useful than the other theorems, but I think enough algebra has been covered for our purposes. Next time I will write a little bit about definitions in topology, a subject I have no more than a week of formal education in, so despite the elementary level it will be a challenge. After that I will try to talk a little about Hilbert’s Nullstellensatz and basis theorem (which I am only marginally more educated in), and then we should be ready to at least attempt a foray into our end goal of algebraic geometry.

Sources:
[1] Dummit and Foote. Abstract Algebra, third edition, pages 97-99. Published 2004.
Used for the statement of isomorphism theorems.