today I want to summarize some basic concepts of Group Theory.
What is a group? Well, I don't mean group as in "bunch of people" or as in "your school drama group". No. I mean it in a technical, mathematical sense. And, as you start seeing it, mathematicians love to be very picky with the details!
This has to do with our previous post on the group, $C_{3v}$, of symmetries of the ammonia molecule $NH_3$. There I mentioned that the six symmetry elements of $C_{3v}$ form a group, and I made the remark that this means something very precise. Here I want to provide you with a very basic introduction to the topic of Group Theory.
I'll try to make this text as self-contained as possible. If you still find things that are not so clear, you may check any elementary book on Algebra, or your class notes. Later I'll provide some links as well.
1.-Definition of Group:
So, what do mathematicians think of when they refer to a group? Here it goes:
- A group consists of two things $(G,\,*)$:
- a collection of elements (we'll denote an arbitrary element by $g$ and the whole collection by $G$)
- and an internal operation $*$ between any pair of elements of $G$, $g$ and $g'$. This means, that the result of "multiplying" both elements is again an element of $G$, i.e., $g* g' \,\in\,G$.
- In addition the pair $(G,\,*)$ must satisfy following properties:
- There is a neutral element $e\in G$ such that
$e* g \,=\,g* e\,=\,g$
for any arbitrary element $g\in G$. - The "product" is associative, i.e.
$(g* g')* g''\,=\,g* (g'* g'')$ - For every element $g$ there is an inverse element, which we'll denote by $g^{-1}$, such that
$g* g^{-1}\,=\,g^{-1}* g\,=\,e$.
As you can see, a group is indeed a collection of "things", somehow. The key point is that we have a way of combining those "things" together and, when we do it, we obtain another thing of that same collection. Furthermore, this combination has to follow some very precise rules (the above three rules) and we you change these rules it won't be a Group anymore; but something else!
Another fun way to express the same thing is saying that those "things" can interact among them and the result of such an interaction is another "thing" of the same kind. That's very much the same as when you write a chemical reaction like, e.g., $H_2\,+\,O_2\,\longrightarrow\,2\,H_2O$: two type of substances (a hydrogen molecule and an oxygen molecule) interact (we symbolize this with the $+$ sign!) to yield another type of substance (water!).
Incidentally, the concept of group shows as well what the essence of Algebra is. Basically, Algebra is about considering a collection of objects and defining one or more internal operations among them. Each time you do so, mathematicians, very pompously say "we defined a structure"! In this case, the structure defined is that of a group. So, what keeps mathematicians so busy about groups, and algebra in general? Once you define a structure, what else is there to do? Don't you then know all about it already? Well, we already know $G$ is a group, yes. However, our goal is to learn about how $G$ is internally structured. Read on to see what I mean.
2.- Types of groups:
In these notes I want to quickly summarize some useful results about groups that should help you understand the things we already talked about and some others to come soon.
In this spirit, I'll start by making a very simple distinction between finite groups and continuos groups. This is already a statement about the internal structure of a group.
The first type are groups that have a finite number of elements, e.g., the group $C_{3v}$ we saw before, which has only $6$ elements. The second type are groups that have an uncountable infinite number of elements, e.g. the group of rotations of the sphere by an angle $\theta\in \left[\right.0,\,2\pi)$ around the the $z$-axis: For every value of $\theta$ there is a rotation $R(\theta)$ which is an element of the group, with $\theta$ being a real number! For instance, the identity is given by $R(\theta = 0)$; another element is a rotation $R(90^\circ)$ of $90^\circ$, etc.
From now on, I will refer to finite groups unless otherwise stated. Although many things will be valid in general, like the definition above, the theory of continuos groups is a world in its own that goes by the name of Lie groups.
3.- The Cayley table of a finite group:
One easy way to summarize a (finite) group $(G,*)$ is by way of writing a table of all possible "products". This is called the Cayley table. (It's obvious that this applies only to finite groups, isn't?)
The Cayley table for our group $C_{3v}$ is as follows (here we use $E$ to denote the identity element $e$):
| $E$ | $C_1$ | $C_2$ | $S_1$ | $S_2$ | $S_3$ | |
| $E$ | $E$ | $C_1$ | $C_2$ | $S_1$ | $S_2$ | $S_3$ |
| $C_1$ | $C_1$ | $C_2$ | $E$ | $S_3$ | $S_1$ | $S_2$ |
| $C_2$ | $C_2$ | $E$ | $C_1$ | $S_2$ | $S_3$ | $S_1$ |
| $S_1$ | $S_1$ | $S_2$ | $S_3$ | $E$ | $C_2$ | $C_2$ |
| $S_2$ | $S_2$ | $S_3$ | $S_1$ | $C_2$ | $E$ | $C_1$ |
| $S_3$ | $S_3$ | $S_1$ | $S_2$ | $C_1$ | $C_1$ | $E$ |
The way we wrote this table is meant to be read as follow: The action of performing first $S_1$ and then $C_1$, i.e., the product $C_1*S_1$ is $S_3$. In other words, the columns specify the right-most element of the product (first operation we apply) and the row the left-most element of the product (second operation we apply).
Notice that this "multiplication" table represents a product that is not commutative: $C_1*S_1\,\neq\,S_1*C_1$!
4.-Definitions:
We'll introduce some basic concepts than have been useful in describing the internal structure of a group.
Given a group $(G,*)$ we will introduce the following definitions:
The order of a group $G$ is the number of elements it contains. It's denoted as $ord(G)$. Thus $ord(C_{3v})=6$.
The order of an element $g\in G$ is defined as the number of times we have to "multiply" $g$ by itself in order to obtain the neutral element $e$.
Exercise 1: The order of $S_1$ is $2$ as it is $S_1^2\,=\,E$. Check the following results: $ord(C_1)=ord(C_2)=3$ and $ord(S_2)=ord(S_3)=2$.
[Remark: For a finite group the order of an element is always defined. Remember that operating two elements of a group always yields another element of the same group. If the group is finite, and you keep "multiplying" $g$ by itself, eventually you must hit $e$! This is not meant as a formal proof of that statement, but rather give an idea of how the proof goes. However, it's trivial to proof by contradiction; just assume it never gives $e$ but a list of, say, four elements $\{g,g^2,g^3,g^4\}$. The reasoning is the same if you consider another number instead of $4$. What happens when you keep calculating powers $g^5,\, g^6,\cdots$ or, say, $g^{3141516}$?]
Subgroup: A group $(H,*)$ is called a subgroup of $(G,*)$ if $H$ is a subset of $G$ and when "multiplying two elements of $H$ yields always again an element of $H$. We can briefly summarize the two conditions as
- $H\subseteq G$
- $\forall h,h'\in H\,;\, h*h'\in H$
Of course, $G$ and $\{E\}$ are always subgroups of $G$. These are the trivial subgroups and we rarely talk about them. We already know $G$ is a group. Our goal is to learn about how it is structured. Keep on reading to learn more about how we can learn more about $G$'s structure. (I know, I said that before already :)
Exercise 2: Find all subgroups of $C_{3v}$.
Give it a try before reading on...
Ok, the subgroups are $\Sigma_1\,=\,\{E,S_1\}$, $\Sigma_2\,=\,\{E,S_2\}$, $\Sigma_3\,=\,\{E,S_3\}$ and $R\,=\,\{E,\, C_1,\, C_2\}$.
That the rotations of the ammonia molecule form a subgroup expresses just the intuitive fact that applying two consecutive rotations yields another rotation, either one of $240^\circ$ ($C_1*C_1=C_2$), one of $120^\circ$ ($C_2*C_2=C_1$) or one of $360^\circ$ ($C_1*C_2 = E$), which is the same as the identity. It is $ord(R)=3$.
Swapping two given hydrogen atoms, say $H_2$ and $H_3$, twice consecutively is the same as not doing anything. Thus, each swapping yields one subgroup. For the swappings $ord(\Sigma_1)=ord(\Sigma_2)=ord(\Sigma_3)=2$.
That's all the concept of subgroup is saying: "Things stay within the given subset of elements"!
The subgroup $R$, taken as a group in itself, is an example of what is called a cyclic group. These are groups that are generated by the repeated multiplication of an element by itself. Such an element is called the generator of the group. $R$ is generated by the rotation $C_1$: $R=\{C_1,\,C_1^2,\,C_1^3\}$. Often, you may as well see that written as \[ \langle C_1 \rangle\,=\,R \] where the left and right angle brackets mean the group generated by the elements in between.
Now you can also understand why mathematicians chose the same word order when talking about a group and an element of a group: The order of an element is the same as the size of the group it generates! Put it simply, order always refers to the size of a group; one just needs to understand which group it refers to.
Exercise 3: Find all cyclic subgroups of $C_{3v}$ and their respective orders.
Here you have, thus, another example of how a group can look like. It could be cyclic, like $R$, or non-cyclic, like $C_{3v}$; it can be finite, like these two examples, or infinite,...And the list of possible features doesn't stop there. Read on.
Cosets: Take a subgroup $H$ of $G$. Take as well an arbitrary element $g$ -it could be an element of $H$; it doesn't matter. We can form the subset of elements constructed as $g*H$, that is, take an arbitrary element $h\in H$ and do the multiplication $g*h$. Do that for all $h$'s and the collection of all products is the subset we denote by $g*H\subseteq G$. This is called the left-coset of the subgroup $H$ through $g$. If we now take another $g'\in G$ and do the same procedure again we'd obtain another different left-coset $g'*H$. We keep repeating this for all $g\in G$ and we obtain the list of all left-cosets of $H$
$H$, $g_1*H$, $g_2*H$, ... , $g_n*H$, ...
We could do the same but doing the multiplication in reverse order and we'll denote this by $H*g$. This is called the right-coset of the subgroup $H$ through $g$. As we did before, this would lead us to the list of all right-cosets of $H$:
$H$, $H*g_1$, $H*g_2$, ... , $H*g_n$, ...
In general, the left and right cosets are different.
Exercise 4: From the Cayley table of $C_{3v}$, we can read the left-cosets of $\Sigma_1$ from its column as
$E*\Sigma_1=\Sigma_1$, $C_1*\Sigma_1=\{C_1,\,S_3\}$, $C_2*\Sigma_1=\{C_2,\,S_2\}$, $S_1*\Sigma_1=\Sigma_1$, $S_2*\Sigma_1=\{S_2,\,C_2\}$, $S_3*\Sigma_1=\{S_3,\,C_1\}$.
In summary, denoting by $C_{3v}*\Sigma_1$ all left-cosets of $\Sigma_1$, we have
\[C_{3v}*\Sigma_1\,=\,\Bigl\{\Sigma_1,\,\{C_1,\,S_3\},\,\{C_2,\,S_2\}\Bigr\}\] Find all the cosets of $C_{3v}$.
Give yourself some time before reading on...
Ok, here are all left and right cosets of $C_{3v}$
\[ C_{3v}*\Sigma_1 = \Bigl\{\Sigma_1,\,\{C_1,\,S_3\},\,\{C_2,\,S_2\}\Bigr\} \\ \Sigma_1*C_{3v} = \Bigl\{\Sigma_1,\,\{C_1,\,S_2\},\,\{C_2,\,S_3\}\Bigr\} \\ C_{3v}*\Sigma_2 = \Bigl\{\Sigma_2,\,\{C_1,\,S_1\},\,\{C_2,\,S_3\}\Bigr\} \\ \Sigma_2*C_{3v} = \Bigl\{\Sigma_2,\,\{C_1,\,S_3\},\,\{C_2,\,S_1\}\Bigr\} \\ C_{3v}*\Sigma_3 = \Bigl\{\Sigma_3,\,\{C_1,\,S_2\},\,\{C_2,\,S_1\}\Bigr\} \\ \Sigma_3*C_{3v} = \Bigl\{\Sigma_3,\,\{C_1,\,S_1\},\,\{C_2,\,S_2\}\Bigr\} \\ C_{3v}*R = \Bigl\{R, \Sigma\Bigr\}\,=\,R*C_{3v}\\ \text{where we have defined} \\ \Sigma = \Bigl\{S_1,\,S_2,\,S_3\Bigr\} \]
Normal (invariant) subgroup: As you can see, for the subgroup $R$ the left and right cosets are the same. When this happens, we call that subgroup a normal subgroup. We can see that $R$ is the only non-trivial normal subgroup of $C_{3v}$.
Symbolically, we wrote the fact that $R$ is a normal subgroup as $C_{3v}*R\,=\,R*C_{3v}$. This looks a lot like the property of commutativity! Any relation to it? Well, in general a group is not commutative. Case in point, our friend $C_{3v}$: For instance, $S_1*C_1=S_2\,\neq\,C_1*S_1=S_3$. However, if we "ignore the details" of which exact pair of elements we are multiplying, but only look at which coset each belongs, in some sense we can recover the commutativity. There is a precise way in which we can "ignore the details". We'll come to that again when we talk about the quotient (or factor) group below.
Why is a normal group also called invariant? This sounds interesting; like there would be something being conserved. Probably, this reminds you of our game CSƆ (Classification, Symmetry, Ɔonservation) , isn't? But, where is there any "transformation" happening? What is there being conserved? Read on; we'll come to it further below.
Wether a given group has or not invariant subgroups, or how many it has depends on each group. This is telling us something about its structure. If the only normal subgroups of $G$ are the trivial ones, we say that $G$ is a simple group. For example, $C_{3v}$ is not a simple group, but $\Sigma_1,\,\Sigma_2\,\Sigma_3$ and $R$ all are simple groups.
Partitions of a group: This is not a topic that you may find elsewhere when talking about groups, but this blog is about making contact with the topic of partitions after all! So, I'll talk about this. :)
A partition of a collection of objects is a division of it into classes such that (1) they all are mutually disjoint and (2) the union of all of them equals the whole collection. Consider, for instance, the left-cosets of $\Sigma_1$,
\[ C_{3v}*\Sigma_1 = \Bigl\{\Sigma_1,\,\{C_1,\,S_3\},\,\{C_2,\,S_2\}\Bigr\} \]
As you can see any two classes are disjoint and their union gives the whole group
\[ \Sigma_1\,\cup\,\{C_1,\,S_3\}\,\cup\,\{C_2,\,S_2\}\,=\,C_{3v} \]
Exercise 4: Check that both the left- and right-cosets of each subgroup form a partition of $C_{3v}$.
\[ \forall g,\,g'\in G\,;\quad g\,\thicksim_{GA}\,g'\,\iff \,g*A\,=\,g'*A \,\Leftrightarrow\,g^{-1}*g'\in A \] That is, two elements, $g_1,g_2$, belong to the same "left-A-class" (left-coset of A) iif the inverse of one by the other one, $g_1^{-1}*g_2$ yield an element of $A$.
Analogously, we can define the equivalence relation given by the right-cosets of $A$, $A*G$. However, in this case the equivalence relation we obtain is, in general, different:
\[ \forall g,\,g'\in G\,;\quad g\,\thicksim_{AG}\,g'\,\iff\,A*g\,=\,A*g' \,\Leftrightarrow\,g'*g^{-1}\in G \]
Notice the change of order in the last step! As the group may not be commutative, both things express different conditions: it could be that $g'*g^{-1}\in A$, but $g^{-1}*g'\notin A$! Case in point, the cosets of, say, $\Sigma_1$: Considering the partition given by $C_{3v}*\Sigma_1$, $C_1$ and $S_3$ belong to the same left-coset, and thus \[ C_1^{-1}*S_3\,=\,C_2*S_3\,=\,S_1\,\in\,\Sigma_1, \\ \text{however,}\\ S_3*C_1^{-1}\,=\,S_3*C_2\,=\,S_2 \notin \Sigma_1 \]
The latter, however, is the condition for belonging to the same right-coset of $\Sigma_1$! Indeed, the equivalent partner of $C_1$ in $\Sigma_1*C_{3v}$ is $S_2$, and we have \[ C_1*S_2^{-1}\,=\,C_1*S_2\,=\,S_1\,\in\,\Sigma_1 \]
Of course, we obtain the same answer if we check the other ways around (it's indeed an equivalence relation, and thus, symmetric!), namely, $S_2*C_1^{-1}=S_2*C_2=S_1\in \Sigma_1$. However, for the right-coset we see again that the check is not satisfied with the inverse on the left: $S_2^{-1}*C_1=S_2*C_1=S_3\notin \Sigma_1$.In summary, when you need to check whether any two elements $g,g'$ belong to the same coset, the test for the left-cosets has the inverse on the left; while that for the right-cosets has the inverse on the right!
Exercise 5: Proof whether or not 1) $C_2$ and $S_3$ belong to the same coset in a) $C_{3v}*\Sigma_1$, b) $\Sigma_2 * C_{3v}$ and c) $C_{3v}*\Sigma_3$.
Exercise 6: Proof that $\thicksim_{GA}$ and $\thicksim_{AG}$, noth are, indeed, an equivalence relation. That is, proof that each is 1) reflexive, 2) symmetric, and 3) transitive. In short, for a generic equivalence relation \[ g\,\thicksim\, g \\ g\,\thicksim\,g'\,\iff\,g'\,\thicksim\,g \\ \text{if}\quad g\,\thicksim\,g'\quad \text{and}\quad g'\,\thicksim\,g''\,\to\,g\,\thicksim\,g'' \] where, $g,g',g''\,\in\,G$ are arbitrary elements of $G$. Demonstrate that both relations satisfy these three conditions.
Don't you wish both checks would give the same relation, that is, would be equally valid? That would make life easier when defining a partition of a group. But this is indeed the case for the cosets of a normal (invariant) subgroup! This explains part of the importance of normal subgroups.
As for a normal subgroup both left- and right-cosets are the same, so are as well both equivalent relations. Thus, a normal subgroup $N$ determines a unique equivalence relation $\thicksim_{N}$ characterize by \[ \forall g,g'\,\in\,G\quad\;\quad g\,\thicksim_{N}\,\iff\,g*g'^{-1}\,\in\,N\,\iff\,g^{-1}*g'\,\in\,N \] We can check this is indeed like that for the case of $C_{3v}$.
Exercise 7: For the normal subgroup $R$ of $C_{3v}$, 1) check that for all pair of elements $g,g'\in C_{3v}$, if $g^{-1}*g'\in R$ then it is also true that $g*g'^{-1}\in R$; 2) check as well that the implication is also true the other way around, that is, if $g*g'^{-1}\in R$, then it is also $g^{-1}*g'\in R$.
Cosets and Action of a Group: You may ask, why is the concept of coset relevant at all? Fine, it's telling us about the so called structure of a group, but, so what? what is this useful for? I'll try to answer this now.
Consider the symmetry transformations that leave invariant the three vertices of an equilateral triangle. Yes, you got it right! These transformations are given as well by our friend $C_{3v}$ (technically, both groups are isomorphic; but we'll come to that later). Indeed, we can swap two vertices or we can rotate the triangle as a whole, with all three vertices simultaneously moving.
Consider now the action of this group, $C_{3v}$ on the vertices. We can represent the three vertices by a $3$-dimensional vector and the group elements by matrices, as we did before. The action of, say, the swapping $S_1$ on any given vertex can be written as \[ \begin{pmatrix}1& 0 & 0 \\0&0&1\\0&1&0\end{pmatrix} \mathbf{e}_i \quad i=1,2,3 \] If right after that transformation, we do another one, say a $120^\circ$ rotation $C_1$, with the present notation, we would write the whole effect as \[ \begin{pmatrix}0&0&1\\1&0&0\\0&2&0\end{pmatrix}\begin{pmatrix}1& 0 & 0 \\0&0&1\\0&1&0\end{pmatrix} \mathbf{e}_i \quad i=1,2,3 \] In other words, the composition of transformations is given by left multiplying with the corresponding matrices.
The group $C_{3v}$ is small. Yet, if I want to understand what it does to the triangle, do we need to remember all matrices of all $6$ representations? do we need to remember all possible combinations of them as well? That doesn't sound like fun. But the left-cosets come here to rescue us!
The cosets are telling us that we just need to consider pair-multiplications (and not products of 3, 4,...100 matrices!) to describe the effect of the group. Even more, for instance, the partition given by $C_{3v}*\Sigma_1$ is telling us that we just need multiplications by either the neutral element $E$ or $S_1$. Indeed, when left-multiplying, $S_3$ is given by $C_1*S_1$ and $S_2$ is given by $C_2*S_1$. That is, we can simply consider the swapping $S_1$ and the rotations, and all three generate all possible transformations!
Mathematically, we can express this by saying that $S_1,\,C_1,\,C_2$ generate the whole group, \[ \langle S_1,\,C_1,\,C_2\rangle\,=\,C_{3v} \]Swap once and rotate the system to your desired orientation! That's all there is to $C_{3v}$! Cool, isn't?
Of course, we may choose differently. If $S_3$ is given by $C_1*S_1$, it is also true that $C_1$ is given by $S_3*S_1$! Analogously for $S_2$. Thus, can also consider just the reflections of $C_{3v}$ to describe all effects of this group! In more mathematical terms, we can therefore also write \[ \langle S_1,\,S_2,\,S_3\rangle\,=\,C_{3v} \]Let's get convinced through an explicit example. Let's say we want to apply the transformation $C_1*S_2*C_2*S_3$. You immediately see that it doesn't start with $S_1$! Actually, there is no $S_1$ at all! But we said that we just need the three reflections, or just one of them and the two rotations! Were we wrong!? Not really. We just need to show that we can rewrite that composition of transformations in terms of the generators we want. For instance, in terms of $\{S_1, C_1, C_2\}$, it is $C_1*(C_2*S_1)*C_2*(C_1*S_1)\,=\,S_1*S_1\,=\,E$; and in terms of $\{S_1,S_2,S_3\}$, it is as well $(S_3*S_1)*S_2*(S_2*S_1)*S_3\,=\,S_3*S_3\,=\,E$.
Given that $C_1$ is the generator of a cyclic group, we can even forget about $C_2$: just repeat $C_1$ twice whenever you see $C_2$! Therefore, for this particular group, it is also true that \[ \langle S_1,\,C_1 \rangle \,=\,C_{3v}.\]This is exactly was is encoded in the fact that 1) $R=\langle C_1\rangle$ and 2) that $R$ is a normal group of $ord(R)=3$. As the whole group $C_{3v}$ has $6$ elements and all cosets of $R$ must have $3$ elements as well, there can only be $2$ cosets! This is the so-called Langrange Theorem.
We could have taken other combinations by considering other left-cosets, and that would give as different choices of transformations with which we could summarize the action of $C_{3v}$ on the equilateral triangle.
I'm leaving something important: What does it mean that any two elements belonging to the same coset are equivalent? Well, it's not the case that wherever you see $C_1$ you can substitute it by, say, $S_3$. So, yes, in what sense are they equivalent? We will however postpone the answer to the next chapter.
I think I'll stop here for now. This post is long enough and way more abstract than what I initially had planned. I will continue with the rest I wanted to talk about in a following post, part II of Group Theory.
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