Tour:Verifying the group axioms
This article adapts material from the main article: verifying the group axioms
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PREREQUISITES: Definition of group. Review Tour:Group if you don't remember the definition clearly.
WHAT YOU NEED TO DO: Read this article carefully, keeping in mind the definition of a group. This should give you a better idea of how to verify whether a set with a given binary operation forms a group. Try applying this procedure with the examples of groups you've seen so far. (If certain techniques mentioned here use ideas you haven't yet encountered, make a note of them and proceed).
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
.
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