Verifying the group axioms

From Groupprops

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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 $G$ .

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

$*:G \times G \to G$

In particular, this means that:

$g * h$ is well-defined for any elements $g,h \in G$
The value of $g * h$ is again an element in $G$

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

Verify associativity

Associativity requires one to pick three arbitrary elements $g,h,k \in G$ , and show that:

$g * (h * k) = (g * h) * k$

There are various strategies for proving this:

If $G$ is a finite set, this may reduce to checking it on all possible triples of elements in $G$
If $*$ is described by means of a mathematical expression, we may be able to simplify the expressions on both sides in terms of variables $g, h, k$ , and show that both sides are equal.
If $G$ is described as a collection of maps from some set $S$ to itself, and the binary operation in $G$ 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 $e \in G$ such that, for all $g \in G$ :

$g * e = e * g = g$

Again, we have some strategies:

If $G$ 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 $g * e = g$
If $G$ is described as a collection of maps from some set $S$ to itself, and the binary operation in $G$ 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 $g \in G$ , there exists $h \in G$ such that:

$g * h = h * g = e$

Again, we have some strategies:

If $G$ 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 $g * h = e$ for $h$ in terms of $g$
If $G$ is described as a collection of maps from some set $S$ to itself, and the binary operation in $G$ is by composition of maps, the inverse of an element is its inverse as a function

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, and it suffices to compute just a left inverse (or just a right inverse).

Subset of a group

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

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

Quotient of a group by an equivalence relation

Suppose $G$ is obtained as the quotient of a group $K$ by an equivalence relation. We want to see whether this equips $G$ 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:

$a \sim b, c \sim d \implies ac \sim bd$

Some worked-out examples

An Abelian group

Here is one example. Consider $G = \mathbb{R} \setminus \{ -1 \}$ and define, for $x,y \in G$ :

$x * y : = x + y + x y$

We want to show that $(G, * )$ is a group.

First, we check the closure of $G$ under $*$ . Namely, we need to check that if $x,y \in G$ then $x * y \in G$ . Suppose not. Then, we have:

$x + y + xy = - 1 \implies (x+1)(y+1) = 0$

which would force either $x = - 1$ or $y = - 1$ , a contradiction to $x,y \in G$ .

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

$(x * y) * z = (x + y + x y) + z + (x + y + x y) z = x + y + z + x y + y z + x z + x y z$

and we also have:

$x * (y * z) = x + (y + z + y z) + x (y + z + y z) = x + y + z + x y + y z + x z + x y z$

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

First, we need to find the identity element. In other words, for any $x \in G$ , we want:

$x * e = x \implies x + e + xe = x \implies e (1 + x) = 0$

Since $x \ne -1$ , we get $e = 0$ .

Finally, we need to compute the inverse map:

$x * y = 0 \implies x + y + xy = 0 \implies y = -\frac{x}{1 + x}$

This gives a formula for the inverse map.

Thus, $(G, * )$ is a group with identity element $0$ and inverse map:

$x \mapsto \frac{-x}{1 + x}$

A group of symmetries

Here's another example. Suppose $S$ is a finite set of points in $\mathbb{R}^3$ . Suppose $G$ is the set of all maps $f: S \to S$ such that for any $x,y \in S$ , the distance between $f (x)$ and $f (y)$ equals the distance between $x$ and $y$ . Define a binary operation in $G$ by composition:

$(f * g)(x) = (f \circ g)(x) = f(g(x))$

We want to show that $(G, * )$ is a group. Note that $G$ is realized as a set of functions under composition.

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

$(f * (g * h))(x) = f ((g * h)(x)) = f (g (h (x)))$

and similarly:

$((f * g) * h)(x) = (f * g)(h (x)) = f (g (h (x)))$

Since this equality holds for every $x \in S$ , we have:

$f * (g * h) = (f * g) * h$

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

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Verifying the group axioms

From Groupprops

Contents

The general procedure

Define the set and binary operation clearly

Verify associativity

Find an identity element

Find an inverse map

In some special cases

When the binary operation is commutative

Subset of a group

Quotient of a group by an equivalence relation

Some worked-out examples

An Abelian group

A group of symmetries

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