 In the previous video, we saw some examples of subgroups and side of larger groups, but one has to be very cautious because when we define a subgroup, you take a subset of the group, and then you argue that, oh, it's a group inside of a group, therefore, it's a subgroup. We have to be careful because in order to be a subgroup, it's not just that we have a subset which has an operation. It has to be a subset using the same operation as the overset in order for it to be considered a subgroup. So let me provide sort of like a counter example to what I'm trying to describe here. So consider the groups Z5 and Z6 with respect to modular addition. So if we think of these as sets, Z5 is just the set 0, 1, 2, 3, 4. Admittedly, these are congruence classes, but we'll just use the representative of 0, 1, 2, 3, 4. So there's really no loss of just thinking of Z5 as these five elements. Z6, on the other hand, would be the set 0, 1, 2, 3, 4, 5, right? And so as sets, this is a very natural subset containment here. Z5 contains everything that Z6 has except it doesn't have the element five itself. But the thing to remember here is that although Z5 can be viewed as a subset of Z6, Z5 by no means is a subgroup of Z6. So we would actually say that Z5 is not a subgroup of Z6 here. And what's wrong here? That, well, Z5 is a subset of Z6 and they're both groups with modular addition, right? The problem is when you talk about modular addition mod five, that is not the same binary operation as addition, modular addition mod six there. And so the removal of five does not make you go from mod six addition to mod five addition. And so the idea is sort of like the following. When if I take the elements one and four in these two sets, take one and four. Well, when you work with Z6, one plus four will be congruent to five mod six and then there's no reduction to do there because we didn't get bigger than six yet. On the other hand, if we switch over and do that same calculation with Z5, you're gonna get one plus five, sorry, one plus four, which is five, yes. But five, or Z5 doesn't actually have a five element. It will reduce down to zero when you do mod six or when you do mod five right here, excuse me. And the key thing to notice here is that five is not congruent to zero when you work mod six, but it is when you work mod five. So these ideas of addition are different operations. It's not the same addition. We might call it modular addition but the addition is dependent on the modulus itself. And so even though it's a subset with modular addition, these are not considered subgroups because the operation in play is inherently a different binary operation. The restriction of addition mod six to the set zero through four does not produce the group Z5 with addition mod five. Now, in comparison though, we could consider the following set. Take the set of just zero, two and four inside of Z6. Now, if we use just these three elements working addition mod six, we can see that we do in fact have a subgroup structure. Notice that zero is the identity. So the set A to need an identity to be the group. So you take the identity of Z6, which is a zero, zero plus two is two, zero plus zero is zero, zero plus four is four, we're good there. And then take two for example, if you take two plus zero, that's a two, two plus two is four and two plus four is zero, zero being six, congruent to six there, right? And so notice that if you operate by two with any element from H, you give back an element in H. And then finally, if you have four, four plus zero is four, four plus two is zero and four plus four is two, you know, eight is congruent to two mod six. And so therefore, if you operate with four, you always give back something in H, zero, two or four. And so the restriction of addition mod six to the set H will give you a binary operation. Operating, you know, adding two elements from H always produces back an element in H. So addition mod six does restrict to an operation on H and therefore H will give us a subgroup. Like I said, it has an operation, right? That operation will be associative because it was already associative in Z6. It has an identity and we also have inverses. Two is the inverse of four and four is the inverse of two. Another example, if what if you take the set zero and three, we could do a similar type of thing. Zero plus zero is zero, zero plus three is three and three plus three is zero. The restriction of addition mod six to this set zero, three gives us a binary operation. It'll be associative because it was already associative in Z6, you have an identity and three is its own inverse. So we have all the group axioms. These are the subgroups of Z6. I mean, of course you also have the improper subgroup Z6 itself and you have also the trivial subgroup just a zero. The important thing though is Z5 cannot be made into a subgroup even if you change the multi, or even if you change the operation, even if you don't use addition mod five anymore, if you switch to addition mod six, you cannot get a subgroup structure out of Z5 because basically they have the following problem. They're gonna have four plus one, which is equal to five mod six, but five is not inside of this set Z5. So we can't make Z5 into a subgroup of Z6. There's some type of restriction there. Topic we'll talk a lot more about in the future.