My abstract algebra class is blowing my mind with its awesomeness, mostly because all the stuff that we're seeing is stuff that we've seen before...but generalised. Why can't we have learned that first?! Well...I guess you walk before you run, don't you?
This post is an introduction (read: definition dump) on groups. As I've just started with this class, please don't hate me if I forget to mention something, or get something pedantically wrong...because seriously, I really don't know what I'm doing. As always, I love getting suggestions, corrections, etc in my ask.
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.
There's some pretty cool results concerning cyclic groups that I have an itch to cover in a future post. As my knowledge stands, I don't think I know enough to really make something super interesting out of it. For now, I just want to notice something awesome: vector spaces are (abelian) groups, if we take all the vectors and their rescalings via scalar multiplication as the set, and vector addition as the binary operation. It's closed under the operation, which is associative and commutative, and it's got the identity and inverses. In fact, the concept of "span" is closely related to cyclic generation of groups. Additionally, while flipping through my Choquet topology book idly, I noticed a section titled "topological vector spaces", which also kicked my brain into action. Vector spaces have a built-in metric function, which could technically automatically induce a topology. From what I can see peeking through the windows, vector spaces are just specialised groups and topological spaces that also come with a whole slew of separate properties. These additional properties are probably what make vector spaces interesting to pull out and study individually. However, following these observations, I can't help but wish I covered topology and algebra before taking linear algebra, even though the opposite order is encouraged at my university. What do y'all think? Leave me an ask!
