# Verifying the group axioms

This is a survey article related to:group

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This survey article deals with the question: *given a set, and a binary operation, how do we verify that the binary operation gives the set a group structure?* This article views the definition of a group as a *checklist* of conditions.

## The general procedure

### Define the set and binary operation clearly

First, identify the set clearly; in other words, have a clear criterion such that any element is either *in* the set or not in the set. For convenience, we'll call the set .

Second, obtain a clear definition for the binary operation. The binary operation is a map:

In particular, this means that:

- is well-defined for
*any*elements - The value of is again an element in

Thus, for instance, the operation which sends real numbers to is not well-defined when is negative and is not an integer; hence, it does not qualify as a *binary operation*.

### Verify associativity

Associativity requires one to pick three arbitrary elements , and show that:

There are various strategies for proving this:

- If is a finite set, this may reduce to checking it on all possible triples of elements in
- If is described by means of a mathematical expression, we may be able to simplify the expressions on both sides in terms of
*variables*, and show that both sides are equal. - If is described as a collection of maps from some set to itself, and the binary operation in is by composition of maps, then associativity is automatic because function composition is associative

### Find an identity element

An identity element (also called *neutral element*)is an element such that, for all :

Again, we have some strategies:

- If is a finite set, this may reduce to checking by inspection.
- If is described by means of a mathematical expression, we may be able to simultaneously solve two generic equations of the form and . Note that we are trying to solve this as an
*equation*in and*identity*in , i.e., it should be true for all . After solving the equation. Note further that if solving just*one*equation already gives a unique solution for , we*still*need to check that that value of works for the other equation as well. - If is described as a collection of maps from some set to itself, and the binary operation in is by composition of maps, the identity element is the identity map

### Find an inverse map

Next, we need to demonstrate that for every element , there exists such that:

Again, we have some strategies:

- If is a finite set, this may reduce to checking by inspection.
- If is described by means of a mathematical expression, we may be able to solve a generic equation of the form for in terms of
- If is described as a collection of maps from some set to itself, and the binary operation in is by composition of maps, the inverse of an element is its inverse as a function

**NOTE**: Due to the somewhat nontrivial fact that monoid where every element is left-invertible equals group, it suffices to actually find a *left* inverse for every group element; if every element has a left inverse, we automatically get a two-sided inverse. Equivalently, it suffices to find a *right* inverse for every group element; if every element has a right inverse, we automatically get a two-sided inverse.

## In some special cases

In some special cases, we can by-pass checking various conditions for being a group. We discuss two special cases here:

### When the binary operation is commutative

When is commutative, then it suffices to find a left identity element (or right identity element), and it suffices to compute just a left inverse (or just a right inverse).

Actually, even for groups in general, it suffices to find just a left inverse, due to the fact that monoid where every element is left-invertible equals group, so we don't really save anything on inverses, but we still make a genuine saving on the identity element checking.

### Subset of a group

Suppose is given to be a subset of a group , and the binary operation on is the restriction to of the multiplication in . Then:

- We need to verify that the binary operation induces a well-defined binary operation in : the product of two elements in is also in .
- We do not need to check associativity of the binary operation, because it holds in
- Instead of trying to
*find*the identity element of , we can simply verify that the identity element in , actually lies inside - Instead of trying to
*compute*the inverse map in , we can simply verify that the inverse map in , sends to within itself.

### Quotient of a group by an equivalence relation

Suppose is obtained as the quotient of a group by an equivalence relation. We want to see whether this equips with the structure of a group. In this case, the only thing we need to check is that the equivalence relation is a congruence. In other words, if is the equivalence relation, we need to check that:

## Some worked-out examples

### An abelian group

Here is one example. Consider and define, for :

We want to show that is a group. (Note: The group is really coming from a concrete realization of the multiplicative formal group law, but we aren't supposed to use that and instead do the proof from first principles).

#### Verifying that the binary operation is well-defined

First, we check the closure of under . Namely, we need to check that if then . Suppose not. Then, we have:

which would force either or , a contradiction to .

#### Associativity

Next, we need to check associativity. We do this using the generic formula. We get:

and we also have:

Now, observe that is commutative (it is symmetric in and ). So it suffices to compute a one-sided identity element and verify the existence of one-sided inverses.

#### Identity element

First, we need to find the identity element. In other words, for any , we want:

Since , we get . Note that , so .

#### Inverse map

Finally, we need to compute the inverse map:

This gives a formula for the inverse map. Note first that the formula makes sense, because , so . Further, the output is not -1, because solving gives , a contradiction. The inverse map is thus a well defined map from to .

Thus, is a group with identity element and inverse map:

### A group of symmetries

Here's another example. Suppose is a finite set of points in . Suppose is the set of all maps such that for any , the distance between and equals the distance between and . Define a binary operation in by composition:

We want to show that is a group. Note that is realized as a set of functions under composition.

- Closure of under follows from the transitivity of the relation of distances being equal.
- Associativity follows from the fact that function composition is associative. Explicitly:

and similarly:

Since this equality holds for every , we have:

- The identity element is the identity map from to . This clearly satisfies the condition for being an element of .
- To show that every map has an inverse, we first observe that any that preserves distances must be injective. That's because if , then the distance between and is zero, so the distance between and is zero, so . Since is a finite set, must be bijective, so it has a unique inverse map. It is clear that this inverse map also preserves distances, so is in .