So last time we did binary structures. Now we want to look at a specific kind of binary structure: namely, groups!

I basically sneakily introduced groups the last time around anyways, so this definition should be quick. A group is a algebraic binary structure that is associative, has an identity element, and has inverses for all its elements. Now we notice that basically everything we're used to is a group: the real/rational/integer numbers with addition, the real/rational/integer numbers (excluding zero) with multiplication... The nice thing about a group is that, given a and b, we can always solve the equation a * x = b. While we may take that property for granted because the stuff we deal with daily are all groups, I guess it's kind of nice to recognise what stellar cooperative algebraic structures they are. An abelian group is a group that's also commutative.

Let's just dive in, shall we? A subgroup of a group (you must have seen this definition coming) is a subset of the group that's still a group under the induced binary operation. As such, we need the subgroup to be closed under the operation, contain the identity, and contain all inverses of its elements. We get associativity fo fwee because the original operation was associative! There's also the parallel subgroup criterion that gives us a more compact (not in terms of sets hee hee hee) way of checking if a subset is a subgroup: the subset has to be nonempty and contain a * b' for all a and b in the subset. Shall we check that being a subgroup and satisfying the subgroup criterion are identical? First, let's say that the subset is a subgroup. Does it satisfy the criterion?

nonempty: yee budday, it's got the identity!

contains a * b' for all a and b in the subset: well, we know that the subgroup needs to contain inverses, so we know that b' is in the subgroup. And subgroups need to be closed under the operation, so fa sho that a * b' is in said subgroup!

Now, the other way around: let's say it satisfies the subgroup criterion. Dawg, be it a subgroup?

associativity: fo fwee!

identity: well we know that the subset isn't empty, so it's got to have an element, x. By the subgroup criterion, x * x' = e is also in the subset (here, we let a = b = x).

inverses: so now we know the subset has got the identity as well as our original element, x. We need to show that x' is in the subset. Ain't no trippin. e * x' = x' is in the subset (here, a = e and b = x).

closure under *: given x and y are in the subset, we need x * y to be in the subset. We know already that the subset has all dem inverses, yo, so we're good to go. Here, a = x, b = y', so x * (y')' = x * y is in the subset.

So we've shown that the subgroup criterion is aptly named: it is a necessary and sufficient condition to show a subset is a subgroup!

So now you might be thinking: man! these group things are kinda cool! how do I get one for myself?! Well, I agree! These groups things are super cool! As for constructing a group yourself...the simplest way is to generate one starting with an element that isn't e. Let's say we pick n, our favourite element, to be in the group. Well, n * n needs to be in the group because of closure. So does n * n * n, and n * n * n * n... basically, all the powers (both negative and positive) of n need to be in the group. And so does e. This kind of group that we got via n is called a cyclic group, and n is called the generator. The shortcut to writing this group is <n>. We notice that we can also take any arbitrary group and find its cyclic subgroups by the same process.