 Welcome back to our lecture series Math 4220, Abstract Algebra 1 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. This is our first video on Lecture 10, which will be based upon contents that can be found in Section 3.3, entitled Subgroups from Tom Judson's Abstract Algebra textbook. And as the name of the section suggests, we're going to talk about subs here, which essentially speaking, a subgroup is going to be a group inside of a group. So imagine we have some group in question here where we're going to have, we'll call the group, the set, G in the operation will be circle here. So suppose this is a group. Now we consider a subset of G here, right? Because when you have a group, the group itself is really just a pair. There's a set with a binary operation on said set that satisfies certain conditions as associativity, identities, and inverses, right? But as the set G is itself a set, it has subsets. And so a subgroup is going to be a group inside of the group. That is, if a subset H is itself a group, we call it a subgroup. Now what does it mean for H to be a subgroup? How does it mean for it to be a group? Well, we have this operation circle here, right? Where circle is an operation from G to G. I should say that if you take two elements of G, we can operate them and get something in G right here. So you take some order pair G1, G2, and this will transform into some element G3. Now, as this is a function, right, it's easy to restrict the domain. So we could restrict the domain of our operation just to BH cross H, right? And so you get H cross H. So if you take the product of any two elements of H, this is still going to give you something, right? But the issue is that the co-domain of the function, right, G, how do we restrict that? Well, that's where it can get a little bit tricky, right? If you just pick some arbitrary sets, the restriction won't necessarily restrict the co-domain. To be a subgroup, we need it so that when you multiply two elements of H together, you get an element in H, right? So you get some third element H, which is in H itself. We know it'll be in G because G is this overgroup that covers it, but we want to make sure that the product of two elements in H is an element in H still. And so that's the idea of a subgroup. The restriction of the operation to H forms an operation on H itself. And we commonly denote that H is a subgroup of G using the symbol H is less than or equal to G in that situation, right? So this little curly one, we'll use to denote subset. So if H is a subset of G, but if it comes to a point, that'll indicate that H is a subgroup. That is, it is a group in its own right. It's a group sitting inside of this larger group. Now, every group there exists at least two subgroups. So the first one is commonly referred to as the improper subgroup, and that's just G itself. Because by definition, if you restrict the operation from G to G, which is really no restriction whatsoever, you'll have this group sitting inside of that. And as such, it's commonly referred to as the improper subgroup. And in contrast, a proper subgroup will be a subgroup over some proper subset of the group. So that's always a guaranteed subgroup, the improper one. Now, if we're interested in proper subgroups, well, you know, we don't have as many guarantees there. There is one guaranteed proper subgroup, and this is called the trivial subgroup, which consists of just the identity element itself. And so let's think about that for a second. Why does that form a group? I should say, why does it form a subgroup? Well, if you take just the identity element, right? And we restrict our attention just to that. Why is this an operation? Well, if you take your identity E, act on E right there, right? Well, since it's the identity, it'll just be the other element, which is itself E, right? Notice that when you take E times E, you're going to get back E. And so this is an example of a binary operation. And so this doesn't form. It does in fact form a subgroup, right? We call this a trivial subgroup. So we have an operation, kachink. This operation is going to be associative. Well, there's really not a lot. There's really not a lot to go on here with associativity, right? You're like, okay, E times E times E, right? No matter how you simplify this, it's just going to be E in the end, right? So associativity is pretty quick, pretty easy. It has the identity because that's the only thing it has. And it has inverses because the inverse of the identity is itself, right? So this does in fact form a group inside of the group. And we call this a trivial subgroup. I should mention that the empty set does not form a subgroup of the group. And the main reason is that the empty set does not contain the identity, right? So there's no way that it could be a group because a group is guaranteed at least one element, the identity. So the empty set cannot be a group. And also I should mention that no group itself can be empty for this same basic reasons. So let me give you some examples of groups that we have seen before with some of their subgroups. So let's take the group of addition with respect to complex numbers. Well, every real number is a complex number. So we can actually see the real numbers as a subset of the complex numbers. With respect to addition, if you take complex addition and restrict it to real number addition, that gives us a subgroup structure. That is the sum of two real numbers is equal to a real number. And the addition of real numbers is the same thing as the addition of complex numbers. What I mean is if we view a real number R as just some real number plus a zero imaginary component, right? So we can visualize any real number as a complex number. Adding together the real numbers is no different than adding together that real number viewed as a complex number. The two additions are the same thing. And therefore restriction of complex addition to real numbers gives you a subgroup structure. Likewise, the rational numbers form a subgroup with respect to addition of the real numbers, right? Every rational number is a real number and the addition of rational numbers is just a special case of addition of real numbers. And so the rational numbers form a subgroup of the real numbers with respect to addition. But likewise, there's a transitivity principle going on here. If Q is a subgroup of R and R is a subgroup of C, that would then imply that Q is a subgroup of C as well. Rational numbers are also complex numbers. And we restrict a complex number just to the numbers, the rational numbers. That addition is the same at meaning. It's the exact same addition and thus we have a subgroup structure. Same thing for the integers, right? Every integer is a rational number, which means it's a real number, which means it's a complex number. And if you take an integer and you add them together, viewing it as an integer, a rational, a real or a complex number, the addition of integers is always the same thing. And therefore additions with respect, I should say integers with respect to addition is a subgroup of the rationales, a subgroup of the reals, subgroup of the complex numbers. And so those are all examples of additive subgroups. Now, if we switch over to multiplication, right, we need to throw out zero because we can't divide by zero. If you take the non-zero complex numbers, the non-zero real numbers is a natural subset of that. The non-zero rationales is a natural subset of that as well. And multiplication of real numbers, it's just a special case of multiplication of complex numbers. And viewing a real number as just a complex number with no imaginary component doesn't change how you do multiplication. R star is a subgroup of C star with respect to multiplication. By similar reasoning, Q star is a subgroup of R star, which then by transitivity is a subgroup of C star. And these are all examples of infinite abelian subgroups. Let me give you some example of finite ones though. Continuing with C star right now, if you take the set plus or minus one and plus or minus I, we can actually argue that this is in fact a subgroup as well. Now, to convince ourselves that it's a subgroup, we might have to consider when we restrict to plus or minus one or plus or minus I, do we have a binary operation, right? And so the Cayley table can be useful in this situation. You'll notice that one is the multiplicative identity here. So we can fill out the table pretty quickly in that regard. And to be a subgroup, right, it has to be a group. It needs an identity. So you're going to need the identity of the ambient group. But then other things, negative one times negative one is one, negative I plus I, this is an abelian group, right? So we can actually just copy the symmetry there to fill the rest of the table. I times I is negative one, I times negative I is one, and then I, negative I times negative I is negative one right here. So you can see that when we restrict complex multiplication from plus or minus one, plus or minus I, this multiplication, we'll take two of these things. This will always produce an element of the form plus or minus one or plus or minus I. So this in fact does form a finite subgroup of order four inside of the multiplication of C star. This is a finite subgroup. And my similar construction, if you just take plus or minus one, this forms a finite subgroup of our star of order two. Clearly you don't get plus or minus I because those are imaginary numbers. Let's look at some non abelian examples, right? So we've learned about the symmetries, the symmetric groups of the symmetries of the set, one, two, three up to N, right? So SN is the set of all permutations of that set. Well, if you think of the dihedral group, remember the dihedral group, this is the set of symmetries of the regular in-gone. We could label, you know, with our in-gone right here, we could label the vertices like one, two, three, four, all the way up to N. And if you look at the possible symmetries of the in-gone, those symmetries are naturally going to be viewed as permutations. So like if you just do rotation, let's rotate counterclockwise by like say by one turn, then one will go to two, two goes to three, three goes to four, all the way up to N goes to one. We can visualize every symmetry of the in-gone as a permutation, and therefore the dihedral group naturally sits inside of the symmetric group. And as they're both permutations, it's the same permutation multiplication there. The dihedral group will be a finite non-Abelian subgroup of SN. So DN always sits inside of SN, where that is the same N, the same number of letters is in play there. Another example we saw, we talked about the Quaternion group before. That was a non-Abelian group of order eight. It consisted of eight two-by-two complex matrices. And the fact, when we argued that Q8 was a group, we essentially argued that Q8 was a subgroup of GL2C, right, the general linear group of two-by-two complex matrices. These are the non-singular two-by-two complex matrices there. Because we had to argue essentially that to be for Q8 to be a group, the restriction of complex matrix multiplication had to form a binary operation on Q8. And we made a similar argument using Cayley tables that in fact, the Quaternion group, it was a group which simultaneously was showing that Q8 was a subgroup of GL2C. And this just gives us some examples of groups that we've learned about already with some of their subgroups.