Verifying the group axioms

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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 $\sim$ is the equivalence relation, we need to check that:

$a\sim b,c\sim d⟹ac\sim bd$

Some worked-out examples

An abelian group

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

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

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:

$x+y+xy=-1⟹(x+1)(y+1)=0$

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:

$(x*y)*z=(x+y+xy)+z+(x+y+xy)z=x+y+z+xy+yz+xz+xyz$

and we also have:

$x*(y*z)=x+(y+z+yz)+x(y+z+yz)=x+y+z+xy+yz+xz+xyz$

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:

$x*e=x⟹x+e+xe=x⟹e(1+x)=0$

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

Finally, we need to compute the inverse map:

$x*y=0⟺x+y+xy=0⟹y=-\frac{x}{1+x}$

This gives a formula for the inverse map. Note first that the formula makes sense, because $x\ne -1$, so $1+x\ne 0$. Further, the output is not -1, because solving $-1=-\frac{x}{1+x}$ gives $1=0$, a contradiction.

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 $G$.