Groups Aren't Abstract Nonsense

I’ve recently been skimming through this book called “Office Hours with a Geometric Group Theorist” which, perhaps unsurprisingly, is about using geometric objects to study groups. It mostly focuses on how group actions on graphs and metric spaces can reveal information about the group1, and contains some pretty nice results. Unfortunately, I have too many in mind for one post, but I would still like to introduce the basic notions of the subject and a few results I enjoyed. I imagine this will be a lengthy post with a mix of introducing theory and neat applications2.

Groups Actions 3

While I didn’t emphasize this much then, towards the beginning of my intro post on group theory, I mentioned that groups often “perform some action on an object.” Here is where we’ll get to see what I meant and why this is useful.

Let $G$ be a group and $X$ a set. A (left) group action of $G$ on $X$ (sometimes denoted $G\curvearrowright X$, pronounced "$G$ acts on $X$") is a function $G\times X\rightarrow X$ where the image of $(g,x)$ is denoted by $g\cdot x$ satisfying
  • $1\cdot x=x$ for all $x\in X$ where $1\in G$ is the identity
  • $g\cdot(h\cdot x)=(gh)\cdot x$ for all $x\in X$ and $g,h\in G$

In essence, a group action is an action of $G$ on $X$ (i.e. a function $G\times X\rightarrow X$) that respects the group structure of $G$. Above, $X$ is just a set so this is all we require. When we later look at actions on graphs or metric spaces, we’ll further require that $G$’s actions preserves $X$’s structure4.

One of the most basic examples of a group action comes from symmetric groups. Given a set $X$, its symmetric group $S_X$ is $S_X=\{f:X\rightarrow X\mid f\text{ bijective}\}$ with composition as the group operation. This has a natural action on $X$; namely $f\cdot x=f(x)$. If you want, you can verify that this gives a group action, but it is pretty tautological. When $X$ is just a set, there’s no additional structure to preserve so we think of its symmetries as just being permutations; hence the name of the above group. When you think about it, for any group action $G\curvearrowright X$, each $g\in G$ induces a function $X\rightarrow X$ and this function turns out to be a permutation. Thus, we have the following Prove that a group action $G\curvearrowright X$ is the same thing as a group homomorphism $G\rightarrow S_X$. If this homomorphism is injective, then we say that the action is faithful. As a corollary to this exercise, we get the following Every group is isomorphic to a subset of a symmetric group. Let $G$ be a group. By the above exercise it suffices to show that $G$ acts faithfully on some set $X$. We can simply take $X=G$ with action given by left multiplication (i.e. $g\cdot h=gh$). This action is faithful since every $g\in G$ acts differently on the identity, so we win. When studying general group actions, there are two basic concepts of extreme importance. Let $G$ be a group acting on a set $X$. Given some $x\in X$, its orbit is Furthermore, its stabalizer (or stabalizer subgroup) is Personally, I like to think of orbits and stabalizers in terms of graphs. Given an action $G\curvearrowright X$, you can form a graph with vertex set $X$, such that for each $(x,g)\in X\times G$ you get an edge from $x$ to $g\cdot x$. In this language, orbits are connected components of this graph and (elements of) stabalizers correspond to self edges. 5

The following two theorems 6 are good to know for general group action knowledge, but I do not believe either will be used in this post, so I won’t bother proving them here. Let $G$ be a group acting on $X$. Fix some $x\in X$. Then, Let $G$ be a finite group acting on a set $X$. For $g\in G$, let $X^g=\{x\in X\mid g\cdot x=x\}$ be the elments fixed by $g$, and let $X/G$ denote the (set-theoretic) quotient of $X$ by the equivalence relation $x\sim y\iff x\in G_y$. Then, To finish this section, I’ll mention a few definitions that should come up in this post. There are a few types of group actions. In the first exercise above, I defined what a faithful action is already; two more of importance are transitive and free actions. A group action $G\actson X$ is called transitive if $G\cdot x=X$ for some (equivalently, all) $x\in X$. The action is free (or $G$ acts freely on $X$) if $\Stab(x)$ is trivial for all $x\in X$ (i.e. no non-trivial element of $G$ fixes any element of $X$) We’ll see that free actions in particular reveal algebraic information about the group involved.


If we’re gonna study groups through their actions, then we’re gonna need objects for them to act on. Sets are all well and good, but are too unrestrictive. The more structure we require on the objects we act on, the more we will have at our disposal to gain information from.

To begin, we will remind the reader of what a graph is, and then introduce how groups act on them. A graph $G=(V,E)$ is pair consisting of a set $V=V(G)$ or vertices and a set $E=E(G)\subseteq V\times V$ of edges. If $(u,v)\in E\iff(v,u)\in E$ for all $u,v\in V$, then we say $G$ is undirected. If $(v,v)\not\in E$ for all $v\in V$, then we say $G$ is simple.

If you don’t already know them, look up definitions for paths, connected graphs, and trees. 7

Graphs are hardly ever written down in terms of explicit vertex and edge sets. More commonly, they are given as some drawing.

Below is an example of an undirected (but not simple) graph

To define a group action on a graph, we need a notion of an (invertible) structure preserving map. This will give us a notion of when two graphs are the same, and the different ways in which we can view a graph as being the same as itself8 will be what we call it’s symmetries. Given two graphs $G,H$, a graph isomorphism is a bijection $f:V(G)\rightarrow V(H)$ s.t. A graph ismorphism $f:V(G)\rightarrow V(G)$ bewteen a graph and itself is called an automorphism. Given a graph $G$, let $\Aut(G)$ denote the set of automorphisms of $G$. Prove that this set forms a group under composition. Let $K_n$ denote the complete graph on $n$ vertices (i.e. every vertex is connected to every other vertex). Then, every vertex is interchangable, so any rearrangement of vertices gives an isomorphism. Hence $\Aut(K_n)\simeq S_n$. $K_{n,m}$ is the graph with vertex set $V=V_1\sqcup V_2$ s.t. $|V_1|=m$, $|V_2|=n$, and $E=V_1\times V_2\cup V_2\times V_1$. e.g. $K_{3,2}$ is pictured below
Calculate $\Aut(K_{n,m})$ when $n\neq m$ and $\Aut(K_{n,n})$ for all $n,m$.
Now that we’re comfortable with graphs and their isomorphisms, we finally define A group action on a graph is a group homomorphism $G\rightarrow\Aut(\Gamma)$ where $G$ is a group and $\Gamma$ is graph. Let $G\actson\Gamma:G\rightarrow\Aut(\Gamma)$ be a group action with $\Gamma$ a graph. Show that this is the same thing as a group action $G\actson V(\Gamma)$ s.t. $(v,u)\in E(\Gamma)\iff(g\cdot v,g\cdot u)\in E(\Gamma)$. If the above exercise seems obvious, then that’s a good sign that we’ve made good definitions. Admittedly, groups acting on graphs don’t really make an appearance in any of the applications I want to talk about today (although they will appear in the “further reading” section at the end), but this is still something one should know. As our only real use of this, we’ll prove a slightly strong version of Cayley’s theorem. Let $G$ be a group with a generating set $S$. Its Cayley Graph (with respect to $S$) is the graph $\Gamma(G,S)=(V,E)$ where $V=G$ and $E=\{(g,gs)\mid\forall g\in G\text{ and }s\in S\}$. The Cayley Graph for $S_3$ with generating set $S={(12),(23)}$ is pictured below (blue edges are $(12)$ and red edges are $(23)$)
Every group is isomoprhic to the automorphism group of some graph.

Fix a group $G$ with a generating set $S$, and let $\Gamma=\Gamma(G,S)$ be its Cayley graph. We will show in particular that $G\simeq\Aut(\Gamma(G,S))$. Now, we clearly have a homomorphism $G\rightarrow\Aut(\Gamma)$ given by $g\cdot v_h=v_{gh}$ where to avoid confusition we denote the vertex set of $\Gamma$ by the symbols $\{v_g\mid g\in G\}$. This action is faithful because, letting $e\in G$ denote the identity, $g\cdot v_e\neq h\cdot v_e$ for $g\neq h$. Hence, we only need to show all automorphisms arise in this way.
To make this part of the proof simple, we'll need to impose one further restriction on our automorphisms: they must preserve edge labels (e.g. $(g,gs)\in E\iff(\phi(g),\phi(g)s)\in E$ for $\phi\in\Aut(\Gamma)$ where $s\in S$ is regarded as the label of that edge). Now, consider an arbitrary $\phi\in\Aut(\Gamma)$ and write $\phi(v_e)=v_g$ where $e$ is the identity. Fix some vertex $v_h\in V(\Gamma)$. We will show that $\phi(v_h)=\phi_g(v_h):=v_{gh}$ by inducting on the length of the shortest (not necessarily directed) path from $v_e$ to $v_h$. This obviously holds when the path has length 0, so suppose the path has length $n>0$, so we can write $h=ws$ where there's a path of length $n-1$ from $v_e$ to $v_w$ and $s\in S$. By our inductive hypothesis, $\phi(v_w)=\phi_g(v_w)=v_{gw}$ so $\phi$ carries the edge $(v_w,v_h)$ to the edge $(v_{gw},v_{gws})$ to $\phi(v_h)=v_{gws}=v_h$ and hence $\phi=\phi_g$, proving the claim.

When inducting in the above proof, we claim that $s\in S$, but in actuality, it’s also possible that $\inv s\in S$ instead. Finish the proof by handling this case.

In the above proof, we assume that automorphisms preserve edge labels. Does the theorem still hold without this extra assumption (hint: 9)?

This realizes every group as the symmetries of some graph. Hence, even the most abstract groups have some kind of concrete realization.

Metric Spaces

We’re almost to the first nice result; I promise. Before we can state it though, I need to introduce the concept of a Metric space and how they’re acted on by groups. Intuitively, a metric space is anywhere you have some notion of distance.

A metric space $(X,d)$ is a set $X$ together with a function $d:X\times X\rightarrow\R_{\ge0}$ satisfying the following:
  • positive definitedness: $d(x,x)\ge0$ for all $x\in X$ with equality iff $x=0$
  • symmetry: $d(x,y)=d(y,x)$ for all $x,y\in X$
  • triangle inequality: $d(x,y)\le d(x,z) + d(z,y)$ for all $x,y,z\in X$
There are plenty of examples of metric spaces already familiar to you
  • Euclidean space $\R^n$ with the Euclidean metric $$d((x_1,\dots,x_n),(y_1,\dots,y_n))=\sqrt{\sum_{i=1}^n|x_i-y_i|^2}$$
  • Euclidean space with the taxicab metric $$d((x_1,\dots,x_n),(y_1,\dots,y_n))=\sum_{i=1}^n|x_i-y_i|$$
  • Any graph $\Gamma$ (really just its vertex set $V(\Gamma)$) with the path metric, where the distance bewteen any two vertices is the length of the shortest path between them.

The next example is of particular importance, and it also a bit surprising at first.

Let $G$ be a group with generating set $S$ and Cayley graph $\Gamma=\Gamma(G,S)$. Then, we can view $G$ also as a metric space with metric given by the path metric on $\Gamma$ (technically, $V(\Gamma)$)!

There’s actually an alternative way to think of this metric 10. Let $\inv S=\{\inv s\mid s\in S\}$ and define the word length of an arbitrary $g\in G$ to be the length of the shortest word in $S\cup\inv S$ that is equal to $g$. Then, we can turn $G$ into a metric space where the distance between $g,h\in G$ is the word length of $\inv gh$.

Show that the word length of $g\in G$ with respect to $S$ is the length of the shortest path from $v_e$ to $v_g$ in $\Gamma(G,S)$. Furthermore, show that this word length construction and the Cayley graph construction give $G$ the same metric (in the sence that $d_\text{word length}(g,h)=d_\text{Cayley}(g,h)$)
Show that the word metric on $\Z^2$ is the same as the taxicab metric.

Naturally, we have a notion of a symmetry of a metric space coming from ways of viewing a metric space as equivalent to itself.

Let $(X,d_X), (Y,d_Y)$ be two metric spaces. An isometry $f:X\rightarrow Y$ is bijective function s.t. $$d_X(x,y)=d_Y(f(x),f(y))$$ for all $x,y\in X$. Furthermore, the set of isometries from $X\rightarrow X$ forms a group denoted $\DeclareMathOperator{\Isom}{Isom}\Isom(X)$.
A group action on a metric space is a group homomorphism $G\rightarrow\Isom(X)$ where $G$ is a group and $X$ a metric space
Give an equivalent formulation of groups acting on metric spaces
  • $\zmod n$ acts on $\R^2$ with the Euclidean metric via rotations
  • $\Z^n$ acts on $\R^n$ with the Euclidean or taxicab metrics by translations
  • $S^4$ acts on a tetrehdron by permuting its vertices
  • Every group acts on itself with the word metric by left multiplication

Finally, let’s see how actions can be used to reveal algebraic information about groups.

First Neat Application

Before I introduce our first real application of these ideas, I need to introduce one more definition.

Let $G$ be a group. An element $g\in G$ is said to be torsion if it has finite order. If $G$ has no (non-trivial) torsion elements, then we say that $G$ is torsion-free

Now, our goal of this section will be to prove the following theorem

If $G$ acts freely on $\R^n$ with the Euclidean metric, then $G$ is torsion-free

Take a moment to let that sink in; from a almost purely geometric situation (a group acting on a metric space), we can make a non-trivial algebraic conclusion. This is a central theme of Geometric Group Theory, and more examples of this will be seen in this post11. For this particular theorem, the proof relies on the following lemma

Let $S\subseteq\R^n$ be some collection of points. A centroid12 of $S$ is a point $w\in\R^n$ minimizing Note that $w$ may not exist if $|S|=\infty$

Any finite set in $\R^n$ has a unique centroid
Let $S=\{\Ith s1,\dots,\Ith sm\}$ be a subset of $\R^n$, and let $f:\R^n\rightarrow\R$ be defined by $f(x)=\sum_{i=1}^nd(x,\ith s)^2$. Then, write $x=(x_1,\dots,x_n)$ and $\ith s=(\ith s_1,\dots,\ith s_n)$, so $$\pderiv f{x_j}=\sum_{i=1}^n2(x_j-\ith s_j)\implies\grad f=2\sum_{i=1}^n(x-\ith s)$$ From this we wee that the unique point the gradient vanishes at is $f=\frac1n\sum_{i=1}^n\ith s$, so this is a critical point. Proving that this is a (global) minimum is left as an exercise. The lemma follows.

With that Lemma, our theorem will actually have a fairly short proof

Let $G$ be a group acting freely on $\R^n$ with the Euclidean metric. Then, $G$ is torsion-free.
Let $g\in G$ be any torsion element, and fix some point $x\in\R^n$. Let $\mc O=\{x,g\cdot x,g^2\cdot x,\dots,g^m\cdot x\}$ be the orbit of $x$ under $\gen g$. Since $\mc O$ is finite, we can apply the lemma to find a centroid $y\in\R^n$. Since $g$ acts by an isometry, $g\cdot\mc O$ must have $g\cdot y$ as a centroid, but $g\cdot\mc O=\mc O$! Hence, $g\cdot y=y$ so $g\in\Stab(y)$. Since we assumed that $G$ acts freely, $g$ must be the identity so $G$ is torsion-free.

Free Groups and Presentations

For the next application I want to cover, we need a little more group theory. Specifically, we need to define free groups, and since it would be a shame to introduce free groups without introducing presentations, we include them here too. The idea behind free groups is that they’re essentially the bare minimum of what’s needed to call something a group; they don’t satisfy any non-trivial relations. We will begin with the definition of a free group, and then give a “categorical” (or “universal”) characterization of them.

Let $S$ be an arbitrary set, and let $\inv S=\{\inv s\mid s\in S\}$ (note: $S\cap\inv S=\emptyset$). The free group on $S$ is the set $F(S)$ of (reduced) words in $S\cup\inv S$ with group operation given by reduced concatnation (e.g. write $ab\inv ba$ as $a^2$). The rank of $F(S)$ is $|S|$ and $|S|=|T|\implies F(S)\simeq F(T)$. The free group on an $n$-element set is denoted $F_n$.
Verify that the above actually defines a group.
$F_1\simeq F(\{a\})$ is the set of words in $a$. Any such word can be written as $a^n$ for some $n\in\Z$ so the map $a^n\mapsto n$ givens an isomorphism $F_1\simeq\Z$
$F_2\simeq F(\{a,b\})$ is the group of words in $\{a,b\}$ so a typical element might look like $a^2\inv bab^{-3}a^3b$. Note that $ab\neq ba$ so $F_2$ is non-abelian. We do have a surjective homomorphism $F_2\rightarrow\Z^2$ given by $a\mapsto(1,0)$ and $b\mapsto(0,1)$

Free groups can take a while to wrap your head around. I remember I used to be enamored with group theory because from so few axioms (only like 3 or 4), you were guaranteed to get an object that was really well-behaved with so much structure, but that all ended when I learned about free groups13. Free groups, and by extension general (non-abelian) groups 14, are trash; they can have too much freedom and/or unintuitive structure. It’s not all bad though; this makes results involving them feel extra interesting.

Construct an explicit embedding $F_3\hookrightarrow F_2$. If that's too easy then construct an embedding $F_4\hookrightarrow F_2$.
The previous exercise shows that you can construct an injection homomorphism $F_2\hookrightarrow F_2$ that is not surjective. Is the opposite possible? Construct an example of a surjective homomorphism $F_2\twoheadrightarrow F_2$ with non-trivial kernel, or prove non exist

This construction of free groups is nice (and necessary), but we could alternatively choose to characterize free groups in terms of a so-called universal property. This has the advantage of including only the defining properties of a free group without tying the definition to any particular construction.

Given a set $S$, we say $F(S)$ is a free group on $S$, if there exists an embedding $S\hookrightarrow F(S)$, and given any group $G$ and set map $S\rightarrow G$, there exists a unique group homomorphism $\phi:F(S)\rightarrow G$ s.t. the following diagram commutes
where the dotted line signifies that this is the map we are claiming existence of.
Show that our previous construction satisfies this criterion
The above characterises the free group on $S$ uniquely up to unique isomorphism
Let $G,H$ be two groups with embeddings $S\hookrightarrow G,H$ satisfying the above criterion. Then, because $G$ is a free group on $S$ and we have a map $S\hookrightarrow H$, this extends to a unique homomorphism $\phi:G\rightarrow H$. Simiarly, we get a unique homomorphism $\psi:H\rightarrow G$ s.t. the left diagram below commutes.
By commutativity of the left diagram, we get that $\psi\circ\phi:G\rightarrow G$ extends the embedding $S\hookrightarrow G$, but the identity function $1_G:G\rightarrow G$ does this as well. Since such a homomorphism is unqiue, we must have $\psi\circ\phi=1_G$. Similar reasoning shows that $\phi\circ\psi=1_H$ so $\phi,\psi$ are isomorphisms.

This universal property of free groups gives a natural segway into our next topic. Intuitively, a presentation of a group is a compact way of writing it down. Instead of specifying every single element and how to multiply them, you only write down some set of generators and relations (e.g. words equivalent to the identity). The notation looks like

so for example, we have $F_1\simeq\gen{a}\simeq\Z$ and $\zmod n\simeq\pres{a}{a^n}$. In order to formalize this, we make use of the following theorem.

Prove that every group is the quotient of a free group

Now, given a group $G$, in order to write down a presentation for it, we first find some free group $F(S)$ and normal subgroup $K\le F(S)$ s.t. $G\simeq F(S)/K$. Then, letting $R\subset K$ be a generating set for $K$, our presentation is

giving a formal definition of the notation15

The dihedral group $D_{2n}$ (symmetries of a regular $n$-gon) has presentation $D_{2n}\simeq\pres{r,f}{r^n,frfr}$ where $r$ is rotation by $2\pi/n$ degrees and $f$ is flipping across the diagonal.
Group presentations are not unqiue. Show that $\pres{x,y}{xyx=yxy}\simeq\pres{a,b}{a^2=b^3}$
$\zmod 2\simeq\pres{a}{a^2}$ is a 2-element group. What is the cardinality of $G\simeq\pres{a,b}{a^2,b^2}$? Can you find a familar group that $G$ contains as a subgroup?

Second Neat Application

Now that we’re a bit more aquainted with how horrible general groups can be, let’s focus on something a bit more familiar

The group of $2\times2$ matrices with integer coordinates and determinant $1$. Linear algebra is a particularly nice subject, and this is a very linear algebraic group, so it must certainly be really nice, right?

$\SL_2(\Z)$ contains $F_2$ has a subgroup

Like I said, general groups are trash. This application will be similar to the last; we’ll prove a lemma that will give us a direct route to this theorem.

Let $G$ be a group generated by two elements $a,b$, and suppose that $G$ acts on a set $X$. If $X$ has disjoint nonempty subsets $X_a$ and $X_b$ s.t. $a^k\cdot X_b\subseteq X_a$ and $b^k\cdot X_a\subseteq X_b$ for all nonzero $k$, then $G\simeq F_2$.
First, convince yourself that every non-empty word in $a,b$ is conjugate to one of the form $g=a^*b^*\dots b^*a^*$ where the stars are arbitrary nonzero exponents. Now, $g$ is not the identity because $g\cdot X_b\subseteq X_a$, so $G$ has no non-trivial relations and the conclusion follows.
Show that every non-empty word in $a,b$ is conjugate to one of the form used in the proof.

The above proof is a gem in and of itself, because it’s so clean. In case some clarity is lost in its brevity, the idea is that as you apply each syllable (i.e. $a^*$ or $b^*$) of $g$ to $X_b$, you keeping moving back between $X_b$ and $X_a$ (like a game of ping-pong), landing away from where you started. This simple lemma will let us prove this section’s main theorem in a rather concrete way.

Fix some integer $m\ge2$. Let $$A =\mat 1m01\text{ and }B=\mat10m1$$ Then, $A,B$ generate a free subgroup of rank 2 in $\SL_2(\Z)$.
It is easily verified that $A,B\in\SL_2(\Z)$ and an induction argument show that $$A^k=\mat1{km}01\text{ and }B^k=\mat10{km}1$$ Now, note that $A,B$ have a natural action on $\R^2$ via multiplication. Consider the sets $$X_A=\brackets{\left.\vvec xy\in\R^2\right||x|>|y|}\text{ and }X_B=\brackets{\left.\vvec xy\in\R^2\right||x|<|y|}$$ Given $v=\hvec xy^T\in X_A$ and $v'=\hvec{x'}{y'}^T\in X_B$, we have $B^kv\in X_B$ since $$|y+kmx|\ge|kmx|-|y|=m|k||x|-|y|\ge2|x|-|y|>|x|$$ for $k\in\Z\sm\{0\}$. Simiarly, $A^kv'\in X_A$ so the ping pong lemma applies.

Results for Future Posts

I mentioned in the beginning that I wouldn’t be able to cover all the results I would like. In this final section, I waat to mention some of things I left out.

If a group acts freely on a tree, then it is a free group.

An immediate application of this lemma16 is the following

Any subgroup of a free group is free

While this fact may seem obvious or benign, it is definitely non-trivial. As an attempt to appreciate this tree lemma, I challenge you to prove this theorem completely algebraically (Spoiler: 17).

Keeping in touch with the theme of free groups, another surprising result is that $\SL_2(\Z)$ actually contains many more free groups than the one I pointed out in the last section.

For all $m\ge3$, the group $$ \SL_2(\Z)[m]:=\ker(\SL_2(\Z)\rightarrow\SL_2(\zmod m)) $$ is free.

This theorem is actually also proved by exhibiting a free action of this group on a tree.

Moving away from free groups, there’s a notion similar to an isometry but weaker called a quasi-isometry.

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A function $f:X\rightarrow Y$ is called a quasi-isometry if there are constants $K\ge1$ and $C\ge0$ s.t. $$\frac1Kd_X(x_1,x_2)-C\le d_Y(f(x_1),f(x_2))\le Kd_X(x_1,x_2)+C$$ for all $x_1,x_2\in X$ and there's a constant $D>0$ so that for every point $y\in Y$, there's some $x\in X$ s.t. $d_Y(f(x),y)\le D$.
If $G$ is quasi-isometric to $\Z^n$ (both with the word metric), then $G$ contains $\Z^n$ as a finite index subgroup.

This theorem is particuarly interesting because it is highly geometric (or at least, the premise is). It is unclear how to even formulate this theorem algebraically if such a thing can be done.

As the name of this section suggests, I may return to give actual proofs of some of these results in a future post, but for now, I think this post has gone on long enough.

  1. I haven’t looked into the book too deeply yet, so maybe this changes towards the end 

  2. Plus many exercises (don’t feel pressured to do them all) 

  3. This section was originally titled “Groups aren’t abstract nonsense” with the title of the post being “Geometric Group Theory”. While writing it, I decided that the concreteness of groups was a recurring-enough theme to be reflected in the title. Secretly though, groups are just groupoids (categories where every morphism is an isomorphism) with one object, so they actually are abstract nonsense. 

  4. This won’t come up here, but it’s worth mentioning that much of representation theory is simply the study of group actions, and specifically, groups acting on vector spaces 

  5. I want to make and insert an explicit example image of this, but was too lazy to do so. Hence, I encourage you to do this yourself with the group D_8 (symmetries of a square) acting on a regular octagon 

  6. Burnside was not the first to prove this lemma. He, himself, attributed it to Frobenius but even before Frobenius, it was known to Cauchy. I do not know if Cauchy was the first to prove it or not. 

  7. Honestly, half of graph theory is just setting up definitions. Also, shameless plus, I defined 2/3 of these in a previous post 

  8. Recall two sets are the same when there’s a bijection between them, and the symmetrices of a set are the bijections to itself 

  9. I don’t actually know the answer to this, but I suspect it doesn’t; at the very least, when trying to prove it without this assumption, I ran into issues showing that edge labels can’t change. If you decide to look for a (small) counterexample, I believe (but do not know for sure) than one should exist among finite groups with two or three generators 

  10. Worth mentioning that we’ve just turned any group into a geometric object: a metric space 

  11. Mostly referenced without proof at the very end 

  12. The book (pg. 40) seems to define a centroid as the minimizer of the sum of distances (not squared distances) 

  13. Interestingly enough, I had the opposite experience with linear algebra. At first I thought abstract vector spaces were cool because they were so abstract and potentially wild. When I first saw that every vector space had a basis, it ruined linear algebra for me because all these strange, crazy things I had been dealing with were essentially just R^n in disguise 

  14. i.e. groups you may not see in your first exposure to group theory. 

  15. To get a group from it’s presentation, just take the free group on its generators and quotient out by the smallest normal subgroup containing all the relations 

  16. Maybe plus the fact that free groups have (infinite) trees for Cayley graphs 

  17. There are ways other ways to prove this, but all the ones I know are geometric 

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