## Axioms of Group Theory

[ This article approximately corresponds to chapter II of the earlier group theory blog. ]

Group theory happens because mathematicians noticed that instead of looking at individual symmetries of an object, it’s far better to take the set of all symmetries. In fact, investigating such a set often reveals deep insights on the structure of the said object. This was the philosophy behind Felix Klein’s ambitious and far-reaching Erlangen program, which was way ahead of its time and rather controversial then.

Back to group theory. Roughly speaking, a group is a set of symmetries. But we’re going to use an axiomatic approach.

Definition. A group is a set G together with a map $G \times G \to G$, given by $(a,b) \mapsto a*b$ such that the following three axioms hold.

(S1) There is an $e\in G$ such that $e*g = g = g*e$ for any $g\in G$. We call this “e” the identity element.

(S2) For any $x, y, z\in G$, we have $(x*y)*z = x*(y*z)$. We say that * is an associative function.

(S3) For any $g\in G$, there is an element $h\in G$ such that $g*h = h*g = e$. We say “h” is the inverse of “g”.

Our language highly suggests that the identity is unique, and that the inverse of each element is unique. Indeed, this is true. Also, associative basically means that if we have a sequence of operations, e.g. a*b*c*d*x, then however we choose to form the brackets (e.g. (a*(b*c))*(d*x) or ((a*b)*(c*d))*x, the resulting element is unchanged. Indeed, by repeatedly applying the associative law we get:

• (a*(b*c))*(d*x) = ((a*(b*c))*d)*x,
• ((a*(b*c))*d)*x = (((a*b)*c)*d)*x,
• (((a*b)*c)*d)*x = ((a*b)*(c*d))*x.

So as long as the order of the terms is unchanged, the result is still the same element of G. For convenience, we sometimes drop the “*” and just write the above as abcdx. However, the order does matter since g*h ≠ h*g for general elements gh of G

Definition. If g*h = h*g for all elements g, h of G, then we say * is commutative. A group where the product * is commutative is called an abelian group.

Note the subtle difference: commutativity refers to the operation * while abelian is an adjective describing the group. In an abelian group, not even the order matters so we can write abcdxabcxdbacxd = … . For convenience, one often denotes an abelian group with addition +, i.e. the above is written as a+b+c+d+x = a+b+c+x+d = ….; also the identity is denoted by 0 and inverse of g is –g.

The reason is purely psychological: mathematicians are accustomed to think of addition as commutative but multiplication as non-commutative in general (e.g. as in matrices). By convention, it’s ok to denote an abelian group operation by ×, but denoting a non-abelian group operation by + will lead to no end of grief.

Basic Notations. Given an element $g \in G$, we define elements $\ldots, g^{-2}, g^{-1}, g^0, g^1, g^2, \ldots$ as follows:

• the element $g*g*\ldots*g$ with n>0 terms is gn;
• g0 is defined to be e;
• g-1 denotes the inverse of g;
• the element $g^{-1}*g^{-1}*\ldots*g^{-1}$ with n>0 terms is g-n.

A bit of checking quickly reveals that the standard rules of exponentiation hold, i.e. $g^{m+n} = g^m * g^n$ for any integers m and n. If we had used addition to denote the operation, then we should write $\ldots, -2g, -g, 0, g, 2g, \ldots$ instead to be consistent.

Comment. Upon one’s first encounter with group theory, one is naturally skeptical: what can we possibly derive from three simple axioms? A priori, it’s not at all clear such an abstract definition has any interesting results. The richness of group theory thus is a pleasantly surprising outcome. We hope the reader will stick with us throughout the ride.

Who first wrote down these axioms? According to recorded history, Cauchy was the first to explicitly define the notion of a group. Earlier, Galois had relied heavily on group-theoretic concepts to prove his famed result that general quintic polynomials over Q cannot be solved via radicals, but because group theory wasn’t available, he had to rely on rather unwieldy notations and constructions.

## Examples Galore

Let’s construct many examples of groups to familiarise ourselves with the concepts. First, the abelian (i.e. commutative) case.

• the set of integers, under addition +;
• the set of rational numbers, under addition +;
• the set Q* of non-zero rational numbers, under multiplication ×;
• the set Z/m of integers modulo m, under addition mod m, where m is a positive integer;
• the set (Z/m)* of integers mod m which are coprime to m, under multiplication mod m.

For example, Z/5 = {0, 1, 2, 3, 4} and (Z/14)* = {1, 3, 5, 9, 11, 13}. The fact that inverses exist in (Z/m)* is a result of Bezout’s identity. Notice that the group operation for Q is addition but that for Q* is multiplication, so it’s inevitable that we sometimes have to use multiplication for the abelian group’s operation. On the other hand, one almost never sees a structure where addition fails to commute.

Next, the non-abelian case.

• the set Sn of all permutations of degree n (i.e. {1, 2, …, n}), under composition (this is called the symmetric group of degree n);
• the set An of all even permutations of degree n, under composition (this is called the alternating group of degree n);
• the set of symmetries of a shape (e.g. square, icosahedron, buckyball), under composition;
• more generally if X is any set, then the set SX of all bijections X → X, under composition;
• the set GLn(R) of invertible real n × n matrices, under matrix product.

Notice that all the above examples of non-abelian groups are really “operations”. For example, permutations are just functions that shuffle indices 1, …, i around. Symmetries are operations that preserve a shape. And matrices correspond to linear functions on a vector space of dimension n. Now it’s easy to see why commutativity fails: a sequence of operations often cannot be arbitrarily permuted around, e.g. you can’t walk through the doorway before opening the door.

And $(ab)^{-1} = b^{-1} a^{-1}$ since you have to undo operations in a reverse order, as the protagonist of Steins Gate realised.

Exercises. Some exercises to test your concept.

1. If G is the set of odd permutations of degree n, is it a group under composition?
2. If G is the set of all permutations of arbitrary degree, is it a group under composition?
3. If G is the set of non-negative integers, is it a group?
4. If R is the set of real numbers, is it a group under the operation x*y := x+y+1?

Answers. Highlight to read.

1. No, because it doesn’t have an identity element.
2. No as it stands, because you can’t compose permutations of different degrees. HOWEVER, you can embed the set G into the group of all permutations of {1, 2, …} so it forms a subgroup, which is something we’ll see in the next chapter.
3. Undefined, since there’s no operation specified.
4. Yes. Check that the three axioms are satisfied, where e=-1 is the identity and -2-x is the inverse of x.

From now onwards, instead of saying G is a group under some specified operation, we’ll just say “G is a group”. Usually, the operation is pretty unambiguous. E.g. if we were to say: consider the group Z/37, the reader should automatically think of addition mod 37. As one delves deeper into higher mathematics, one learns to simplify notation to preserve one’s sanity.

## Order of Element/Group

The order of a group G refers to the number of its elements. This is usually denoted #G or |G|. We can have |G| = ∞. For example, |Z| = ∞, |Z/m| = m and |(Z/m)*| = φ(m), which is the number of modulo classes mod m which are coprime to m. Also, |An| = n!/2 for n > 1, since product with (1, 2) gives a bijection between the set of odd permutations and the set of even permutations.

On the other hand, the order of an element g of G, is the smallest positive integer m such that $g^m = e$. If no such m exists, then the order is said to be infinite. The notation for order is o(g) or |g|, e.g. in (Z/48)*, we have o(5) = 4, [ Writing |5| = 4 would have been a notational nightmare! ]

Now if $a, b \in G$ are of finite order, say $a^m = b^n = e$, then it is not necessarily true that ab is of finite order. However, if G is abelian, then since we can swap around the order of ab, we get

$(ab)^{mn} = a^{mn} * b^{mn} = e*e = e$.

Conclusion: in an abelian group, the product of two elements of finite order is also of finite order.

Challenge : find a group G with elements ab such that $a^2 = b^2 = e$ but ab is of infinite order. [ Sample answer (highlight to read): consider the group G of all permutations of the set of integers Z. We have functions f(x) = –x and g(x) = 1-x. Then f(f(x)) = g(g(x)) = x but the composition gf(x) = 1+x clearly has infinite order. ]

## Isomorphisms

Let G and H be groups. An isomorphism is a bijective map f→ H such that f(xy) = f(x)f(y) for all elements xy of G. If such an f exists, we say G and H are isomorphic. Basically, this just means that G and H are structurally identical and one can get from one group to the other by a mere relabeling of elements.

For example, Z/3 = {0, 1, 2} and the alternating group A3 = {e, (1, 2, 3), (1, 3, 2)} are isomorphic. It’s quite easy to find all isomorphisms, namely fZ/3 → A3 must take 0 to e, and either [f(1) = (1,2,3), f(2) = (1,3,2)]  or  [f(1) = (1,3,2), f(2) = (1,2,3)].

So there are two possible candidates for the isomorphism f.

We shall now expound on this seemingly trivial point in excruciating detail. Please bear with us.

To reiterate, there’re two possibilities for an isomorphism between groups Z/3 and A3. Furthermore, one isomorphism is as good as the other. Mathematicians say that the two groups are isomorphic but not canonically isomorphic. [ Warning: failure to understand canonical isomorphisms can cause much pain when one ventures into deeper mathematics, so we might as well explain it as thoroughly as we can now. ]

If two groups (or any structures) are said to be canonically isomorphic, it doesn’t mean that there’s a unique isomorphism between them. It just means there’s an unambiguously “right” isomorphism between them which is pleasing to one’s mathematical sensibilities. The whole point is that if we have three different structures A, B and C, then the presence of “canonical” isomorphisms $\phi : A \to B$, $\psi : B \to C$, $\theta : A \to C$, must imply that $\psi\phi = \theta$.

More generally, when we deal with multiple structures, we must make sure that the resulting function is “path-independent” in the diagram of maps. Mathematicians just say that the diagram commutes. In short, if we write out the algebraic objects and the isomorphisms between them, being canonical usually forces the entire diagram to commute since all are defined in a “natural” manner.

For example, if V is a finite-dimensional real vector space, then the dual V* is defined to be the space of linear maps V → R. Since V and V* have the same dimensions, they’re isomorphic. But there’s no natural way to define an isomorphism without exploiting the structure of V. On the other hand, V and V** are canonically isomorphic, so one often hears mathematicians / physicists saying “the dual of dual of a vector space is itself”.

For another example, the groups Z/6 and (Z/9)* are isomorphic but not canonically so. On the other hand, the group Z/15 is generally considered to be canonically isomorphic to Z/3 × Z/5. Here, (x mod 15) gives the pair (x mod 3, x mod 5); the fact that this is an isomorphism follows from elementary number theory. Likewise, (Z/21)* is generally considered to be canonically isomorphic to (Z/3)* × (Z/7)*.

For yet another example, if G and H are groups, then the set-theoretic product G × H has a group structure by defining (gh) * (g’h’) = (g*g’h*h’). Then the group G × H is canonically isomorphic to H × G since there’s a natural map $G\times H \to H \times G$ which takes (gh) to (hg). Likewise, the triple products (G × H) × K and G × (H × K) are canonically isomorphic, so one can say that taking group products is a commutative and associative operation.

We’ll just leave it at that for now, but it’s hard to over-emphasize the importance of thinking in terms of canonical maps.

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