# Structure of Automorphisms

This post will be on the higher-end of things, mostly because I had just learned this and really wanted to get the opportunity to share it.

Today while studying Algebra I learned of a group in a very odd place in my opinion. It’s not too reliant on too much — little enough for me to be alright with defining everything from the ground-up. If you haven’t had Category Theory before, this’ll be a hellish ride, but hopefully you’ll enjoy it.

Our goal is to find a group in the midst of the abstract world of Category Theory. So we have a picture of what we’re hunting down, let’s define a group now.

Definition 1.1: The nonempty set $G$, endowed with the binary operation $\ast$, (briefly, written $(G, \ast)$, or simply $G$ if the operation can be understood) is a group if

1. the group is closed under $\ast$, that is,

$(\forall g, h \in G): g \ast h \in G$;

1. the operation $\ast$ is associative, that is,

$(\forall g, h, j \in G): (g \ast h) \ast k = g \ast (h \ast k)$;

2. there exists an identity element $e_G$ for $\ast$, that is,

$(\exists e_G \in G)(\forall g \in G): g \ast e_g = e_g \ast g = g$;

3. every element in $G$ has an inverse with respect to $\ast$, that is,

$(\forall g \in G)(\exists h \in G): g \ast h = e_G = h \ast g$.

Example 1.2: You’ve actually encountered groups in a lot of places, albeit if this is your first time hearing about them, you might not immediately realize so. For instance, $(\mathbb{R}, +)$ is a group! If you take any three real numbers $a,b,c \in \mathbb{R}$, then most definitely $(a+b)+c=a+(b+c)$. What about the identity? Well, $e_\mathbb{R} = 0$! Because we have $a+0=0+a=a$, just as needed. And does our identity satisfy that last property? Yup, if you have a real number $a$, then there exists another real number, $-a$, where $a+(-a)=(-a)+a=0$.

Some other groups (which you should check satisfy the properties on your own!) are $(\mathbb{C}, +), (\mathbb{Z}, +),$ and $(\mathbb{Z}_p, \cdot_p)$ where $p$ is prime and $\cdot_p$ is multiplication modulo $p$ (try checking when $p$ is not prime — notice something strange? Try and prove that strange fact you uncovered!)

Anyways, now that everyone knows what a group is, let’s go somewhere that seems wildly different. Let’s jump into some abstract nonsense now, shall we?

Definition 1.3: A category C consists of

• a class $\text{Obj(\textbf{C})}$ of objects of the category; and
• for every two objects A, B of C, a set $\text{Hom}_\textbf{C}(A,B)$ of morphisms with the following properties

1.  Identity elements exist: for every object A of C, there exists (at least) one morphism $\text{1}_A \in \text{Hom}_\textbf{C}(A,A)$, the ‘identity’ on A.

2. One can compose two morphisms: two morphisms $f \in \text{Hom}_\textbf{C}(A,B)$ and $g \in \text{Hom}_\textbf{C}(B,C)$ determine a morphism $gf \in \text{Hom}_\textbf{C}(A,C)$.

3. The ‘composition law’ is associative: if  $f \in \text{Hom}_\textbf{C}(A,B)$, $g \in \text{Hom}_\textbf{C}(B,C)$, and $h \in \text{Hom}_\textbf{C}(C,D)$, then $(hg)f=h(gf)$.

4. The identity morphisms are identities with respect to composition: that is, for all $f \in \text{Hom}_\textbf{C}(A,B)$ we have $f \text{1}_A = f$ and $\text{1}_B f = f$.

5. Lastly, the sets $\text{Hom}_\textbf{C}(A,B)$ and $\text{Hom}_\textbf{C}(C,D)$ must be disjoint unless $A=C$ and $B=D.$

Quickly side notes, a morphism $f \in \text{Hom}_\textbf{C}(A,B)$ for instance is sometimes just shortened $f \in \text{Hom}(A,B)$ if category is known, or even just $f: A \to B$. Also, $\text{Hom}_\textbf{C}(A,A)$ is called the set of endomorphisms of A and may be written $\text{End}_\textbf{C}(A)$.

I won’t lie, that was a lot to swallow at once. To sort of get a better feel for categories, let’s take a look at one we’re well accustomed to.

Example 1.4: The category Set consists of

• $\text{Obj(\textbf{Set})} =$ class of all sets
• For A, B in $\text{Obj(\textbf{Set})}$ (that is, they are sets) we have $\text{Hom}_\textbf{Set}(A,B) = B^A$, where $B^A$ denotes the set of all functions $f: A \to B$.

Good further assurance you understand categories would be to go and verify that Set follows the axioms stated for categories in Definition 1.3. Of course the identity for some set A is the identity function $\text{id}_A: A \to A$, where $\text{id}_A (a)=a$, and composition of morphisms is regular composition of functions.

Oh, quick side note, the reason all the functions $f: A \to B$ are denoted $B^A$ is because the cardinality of the set is $|B|^{|A|}$. Try and prove this for yourself!

You can find a lot more examples of categories online, but for now, Set should be enough to provide understanding of them.

Definition 1.5: Let be a category. A morphism $f \in \text{Hom}_\textbf{C}(A,B)$ is an isomorphism if it has a two-sided inverse under composition: that is, if $\exists g \in \text{Hom}_\textbf{C}(B,A)$ such that $gf = \text{1}_A$ and $fg = \text{1}_B$.

Proposition 1.6: The inverse of an isomorphism is unique.

Proof: Let’s assume an isomorphism $f: A \to B$ has two inverses $g_1$ and $g_2$. The standard trick is to compose on the left by one morphism, and on the right by the other, then apply associativity:

$g_1 = g_1 \text{1}_B = g_1 (fg_2) = (g_1 f) g_2 = \text{1}_A g_2 = g_2$.

Since the inverse of an isomorphism $f$ is unique, there’s no harm in denoting it as $f^{-1}$.

Proposition 1.7: With notation as above:

• Each identity $\text{1}_A$ is an isomorphism and its own inverse.
• If $f$ is an isomorphism, then $f^{-1}$ is an isomorphism, and further $\left( f^{-1} \right) ^{-1} = f$.
• If $f \in \text{Hom}_\textbf{C}(A,B)$$g \in \text{Hom}_\textbf{C}(B,C)$ are isomorphisms, then the composition $gf$ is an isomorphism and $(gf)^{-1} = f^{-1} g^{-1}$.

Proof: All these tend to follow immediately from their definitions. For the third point for instance, we could verify it’s a left-inverse (and the right-inverse verification is the same flavor as this):

$(f^{-1}g^{-1})(gf)=f^{-1}(g^{-1}g)f=f^{-1}(\text{1}_B f) =f^{-1}f=\text{1}_A.$

We’ve made it to our final goal. Automorphisms.

Definition 1.8: An automorphism of an object A of a category C is an isomorphism from A to itself. The set of automorphisms of A is denoted $\text{Aut}_\textbf{C} (A)$; it is a subset of $\text{End}_\textbf{C}(A)$.

By Proposition 1.7, composition grants $\text{Aut}_\textbf{C} (A)$ a very familiar structure:

• the composition of two elements $f, g \in \text{Aut}_\textbf{C} (A)$ is an element $gf \in \text{Aut}_\textbf{C} (A)$;
• composition is associative;
• $\text{Aut}_\textbf{C} (A)$ contains the element $\text{1}_A$, which is an identity for composition (that is, $f \text{1}_A = \text{1}_A f = f$);
• every element $f \in \text{Aut}_\textbf{C} (A)$ has an inverse $f^{-1} \in \text{Aut}_\textbf{C} (A)$.

Can you recognize things? $( \text{Aut}_\textbf{C} (A), \circ)$, where A is any object in a category C and $\circ$ is the definition composition in that category forms a group! There’s the structure underlying automorphisms. In the midst of one of the most “abstract” subjects of mathematics, still lies structure.

It’s almost like you expected it, right? Not just because I conveniently defined groups before all of this, but that morphisms in categories had similar properties to groups. They had associativity, identities, all that lacked was inverses and closure — and that’s what the properties of isomorphisms provided.

Example 1.9: Let’s take a quick glance at the group $( \text{Aut}_\textbf{Set} (A), \circ)$. We have $\text{End}_\textbf{Set} (A) = A^A$, that is, it is the set of all functions $f: A \to A$, including strictly injective, neither injective nor surjective, etc. Our subset $\text{Aut}_\textbf{Set} (A)$ of $\text{End}_\textbf{Set} (A)$ consists of only the isomorphic functions (bijections!) $f: A \overset{\sim}{\rightarrow} A$. Our group has elements that are bijective functions and the binary operation is function composition! It should be straightforward showing this group adheres to the group axioms.

Now the coolest part in this isn’t that we discovered a group within the category Set — it’s that we proved that in any arbitrary category there exists a group $( \text{Aut}_\textbf{C} (A), \circ)$! And we did that working with abstract concepts.

Automorphisms will revisited again the future, so try to get a grasp on them early! 🙂

Lastly, some sources for anyone who wants to read more:

1. I highly recommend checking out Group Theory and the Rubik’s Cube by Janet Chen. It’s not a book, just a 39-page PDF, “notes are based on a 2-week course that I taught for high school students at the Texas State Honors Summer Math Camp.” It’s very fun read and includes exercises to ensure your understanding. 🙂
2. Algebra: Chapter 0 by Paolo Aluffi is a beautiful introduction to Abstract Algebra with a good reliance on Category Theory. It’s what I’m currently going through and is undoubtedly one of my favorite texts — Aluffi is such a funny and light-hearted author.

The next post won’t be so high-up, I plan on covering some Cryptography, so that should be worthwhile. See you then!