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 $a \in G$:

$a * e = e * a = a$

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 simultaneously solve two generic equations of the form $a * e = a$ and $e * a = a$. Note that we are trying to solve this as an equation in $e$ and identity in $a$, i.e., it should be true for all $a \in G$. After solving the equation. Note further that if solving just one equation already gives a unique solution for $e$, we still need to check that that value of $e$ works for the other equation as well.
• 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 $a \in G$, there exists $b \in G$ such that:

$a * b = b * a = e$

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 $a* b = e$ for $b$ in terms of $a$
• 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

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 $G$ 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:

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

Some worked-out examples

An abelian group

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

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

We want to show that $(G,*)$ 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 $G$ 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 $x,y \in G$.

Associativity

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.

Identity element

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

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

Since $x \ne -1$, we get $e = 0$. Note that $e \ne -1$, so $e \in G$.

Inverse map

Finally, we need to compute the inverse map:

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

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. The inverse map is thus a well defined map from $G$ to $G$.

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 $G$.