For reference, here’s the definition of a group again:
- a set
- a special element
- a binary operation on
- is associative, that is, whenever , , and are elements of
- is the identity for , that is, for every
- For every element , there is another element , called the “inverse” of , such that .
Today let’s just look at some examples (and non-examples).
We’ve already seen one example from last time: take to be the set of all integers , with the operation of addition. Adding two integers gives us another integer; addition is associative; and it has as an identity. For any , its negation is the inverse: .
On the other hand, the integers with the operation of multiplication do not form a group. Multiplication is associative and has an identity (namely, ), but in general there are no inverses. The inverse of, say, would be some other integer such that , which does not exist.
Motivated by the above non-example, we can think about the set of all rational numbers with the operation of multiplication. Does this form a group? Multiplying two rational numbers yields a rational again; multiplication is associative; and it has as an identity. Given some its inverse is since . Right? Well, not so fast! The problem is that every rational number has a multiplicative inverse except zero. (If then , and is not well-defined.) So with the operation of multiplication is very close to being a group but not quite!
We can fix this by simply excluding zero: is the set of all nonzero rational numbers. This does indeed form a group with the operation of multiplication: multiplying two nonzero rational numbers yields another nonzero rational; is a nonzero rational which serves as the identity; and any nonzero rational has (another well-defined nonzero rational) as its inverse.
Consider the set which has elements, consisting of the first nonnegative integers. Define the operation which works by adding and then taking the remainder . For example, , but , since . In other words, addition “wraps around” the end of the set. Does this form a group? The operation always gives us another element of as its output; it is associative; and is the identity. What about inverses? Well, the inverse of is , since . For example, , , and so on. So this does make a well-defined group (it is sometimes called the cyclic group of order ).
A group can’t have zero elements, since there has to be an identity element. Can we have a group with one element? Sure we can (it is called the trivial group). Let and define the binary operation as . You can check that this is associative and has as its identity, and that is its own inverse. (You can also check that this is super boring.)
How about a group with two elements? Let and suppose is the identity for the binary operation. That means and . So the only thing left is to define what is. It might seem that we have two choices, but in fact we are not allowed to define , because then would not have an inverse: if , there would be nothing we can multiply by in order to to get . So we are forced to choose . But from there we can check that all the group properties do in fact hold. Note we have just proved that there is essentially only one group of size two (I say “essentially” because you could always call the elements something else, but that doesn’t really change the structure of the group). We saw above that is a group for any (actually, for ), so must be the same as the we just defined, and indeed it is: is and is . , just like we defined .
You might like to think about groups with or elements. How many different possibilities are there?