 Welcome back to our lecture series Math 42-20, abstract algebra one for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misseldine. In lecture 33, at the end of it, we prove the so-called first isomorphism theorem for groups. In lecture 34, I want to prove the second and third isomorphism theorems, for which before we can prove them, there's a lima that we need to prove, which honestly this result is fairly interesting in its own right. We'll actually call it a proposition, upgrade it a little bit. This has to do with when is the product of subgroups, actually a subgroup. Let's say that G is a group, finite or infinite doesn't matter. We have a subgroup H and a normal subgroup N. Both H and N are subgroups, they'll be closed under multiplication, identity, and inverses, but N will also be closed under conjugation since it's a normal subgroup. With these assumptions in hand, if we take the product that is the Frobenius product of N times H times N, we mean that we're gonna take every possible product of something from H and something from N that gives you a set. Then under these circumstances where H is a subgroup and N is a normal subgroup, the product H times N is a subgroup of G. And in fact, if N, excuse me, if H is a normal subgroup of G, then the product HN is in fact also a normal subgroup of G. So if you take a product of two subgroups and one of them is normal, then the product will be a subgroup. And if they're both normal, then the product will be a normal subgroup. Now I should mention that these are sufficient conditions to guarantee when the product of two subgroups of a group will be a subgroup or a normal subgroup. This is not a necessary condition. There are conditions for which you can take a product of a non-normal subgroup or the non-normal subgroup, and that is the product could be a subgroup or even a normal subgroup, in fact. So we won't explore every possibility, but let's prove this very important result right here. So we're gonna first show that we're only gonna assume H is a subgroup for a moment and that N is normal subgroup. Let's prove that H times N is a subgroup. So it's gotta be closed under the identity, closed under multiplication, closed under N versus we'll proceed in that manner. So let's first show that HN has the identity. This is pretty easy. Since H is a subgroup, it'll contain the identity. Let's call it E, the identity of G. Likewise, since N is a subgroup, it'll also contain E. And so because E is in both H and N, E times E will be in HN, but as this is the identity, it's actually idempotent and this is just equal to E. So HN contains the identity element E, first thing checked. Now let's show that it's closed under multiplication. Suppose we have two elements, we'll call them X and Y, which live inside the set HN. We wanna show that X times Y is contained inside of HN. Well, since X belongs to HN, that means there's some element little H inside of H and some little N inside of capital N so that X equals HN. And likewise, since Y is inside of HN, there's some H prime in H and some N prime in N so that Y equals H prime N prime. Being inside the set HN guarantees these factorizations of the elements X and Y. All right, and so this is why it's gonna be important that we're normal. So remember, if you're a normal subgroup, that means all of your left cosets are actually equal to their corresponding right cosets. So for example, the coset H prime N is equal to N H prime. What this means for us is that whenever you have an element of N and you multiply it by the element H prime, you can commute H prime past the N, although you might switch to a different N inside of the set. So in particular, if you take the product N times H prime, H prime can commute with N prime, but there's a price, you're gonna switch from little N to a different element of the subgroup capital N. Let's call that element N double prime. So N double prime is inside of N right there. So we can move H prime past N by switching it to some other element of the subgroup N. That's what's guaranteed by normality because left and right cosets are the same thing. All right, so now let's look at closure here under multiplication. If we take the product X times Y, well, X has the factorization H N, Y has the factorization H prime N prime, which as we're in a group, I can just re-associate these things. So let's put the parentheses around the N and H prime just to focus on that for a moment. Well, the normality of the subgroup N, like we said earlier, says that N H prime is equal to H prime N double prime for some potentially different element of N. So N H prime becomes H prime in double prime. Redoing the parentheses again, you get an H times H prime and N double prime times N prime. Well, as H is a subgroup, if you take the product of two things in H, that'll be inside of H. So H times H prime is inside of H. Likewise, N is a subgroup. So if I take the product of two things in N, N double prime and N prime, you'll get something in N. So this looks like, oh, this belongs to H, this belongs to N. So the product belongs to HN right there. So therefore, we see that HN is closed under multiplication. The last thing to show that it's a subgroup, we need to show inverses. Again, since N is a normal subgroup, the left coset H inverse N is equal to the right coset H, excuse me, N H inverse. So again, what does it mean for these two cosets to be the same? It means that if I take the element N double prime, excuse me, N inverse, which belongs to N, and you multiply that by H inverse, I can bring H inverse to the side, I can commute it with N inverse, but I'm gonna switch to a different element of N. Let's call that element N triple prime. So we get this factorization, or more of a commutation I should say, N inverse H inverse will equal H inverse N triple prime, where N triple prime could be N prime, or N inverse all we know, but it's just some element of N, potentially different, very likely different. And so then when you look at X inverse, well, X remember is HN. If we take HN inverse by the shoe sock principle, we've switched the order of these things. You put your shoes on, then, or I should say you should put your socks on, then your shoes, then you take your shoes off, then you take off your socks. So it switches things around when you take the inverse. So you have N inverse H inverse. We want it to be an H element and then an N element. So it's kind of backwards. This is where this commutation principle comes into play. N inverse H inverse is equal to H inverse N triple prime. Well, H inverse is inside of H because H is a subgroup, and N triple prime is inside of N. So this belongs to HN, and therefore HN is closed under inverses. This then shows that HN is a subgroup of G. So far we've done, we assume that H was a subgroup, and we assumed that N was a normal subgroup. We got that H times N was a subgroup. Now, let's suppose that H is a normal subgroup of G. We will continue to believe that N is a normal subgroup of G. Let's now show that H times N is closed under conjugation, thus showing that it's a normal subgroup. So if we take an arbitrary element G inside of the group G, and let's take our element X here, which again, X is an arbitrary element of H times N. So let's look at the conjugate of X by G here. Well, since X belongs to HN, it can be factored as some little H times some little N, which we can then insert between H and N, some identity element E, for which that identity element is gonna be G inverse G that we can see right here. Then redoing parentheses, of course, we then get GHG inverse, and then GNG inverse. The significance of this, of course, is that when you look at this element GHG inverse, this is a conjugate of H. As H belongs to H, little H belongs to big H, its conjugate will belong to H because it's normal. Same thing here. We're looking at a conjugate of little N, little N belongs to big N, so therefore its conjugates will belong to big N. We have something from H times something from N that gives you something in HN. So therefore, it's inside of the set, and this shows us that HN is closed under conjugation, for which that then shows us that HN is a normal subgroup of G.