MATH 370 - EXAMPLES: GROUPS, SUBGROUPS, COSETS DR. ZACHARY SCHERR There seemed to be a lot of confusion centering around cosets and subgroups generated by elements. The purpose of this document is to supply some examples for you to look at to get comfortable with these ideas. We ve seen many of these examples in class already, but I felt it would be helpful to put them all into a document. 1. Cyclic Groups 1.1. Z. The group G = Z under addition is an infinite cyclic group. This is because it is generated by the element 1. In our notation, this says Z = (1). Of course other elements may generate the same group, for example we also have Z = ( 1). Let s look at some nontrivial subgroups. Let H = (5) G. Then H is a cylic subgroup since it is generated by a single element of G. The subgroup H consists of all multiplies of 5, and so it sometimes written as (5) = 5Z. What are the right cosets of H in G? We can just start listing them until we ve exhausted all elements of G. We have H + 0 = {..., 10, 5, 0, 5, 10,...} H + 1 = {..., 9, 4, 1, 6, 11,...} H + 2 = {..., 8, 3, 2, 7, 12,...} H + 3 = {..., 7, 2, 3, 8, 13,...} H + 4 = {..., 6, 1, 4, 9, 14,...} Here we see that every element of G is in one of these five cosets, and these cosets are all disjoint. Thus the index of H in G is 5. We ve already proved in chapter 1, albeit in a different language, that in fact the index of H = (n) in G is n. Notice that I used the absolute value since, for example, the subgroup (5) is the same as the subgroup ( 5). We can look at subgroups generated by any number of elements of Z. For example, which subgroup is H = (2, 3)? Again, this subgroup consists of all elements of Z which are combinations of 2, 3, 2 and 3. It s plain to see that 1 H since 1 = 3 2, but then this must mean that H = G. The reason is simple, the element 1 generates all of G = Z, and the second we know that 1 H we must get every element of G in H. Thus G H but 1
2 DR. ZACHARY SCHERR also H G and so G = H. You can try this with other numbers. Check, for example, that H = (6, 9) is really the same subgroup as (3). Again, we ve proved this before but in a different language. You can look at subgroups generated by any number of elements. Here s an interesting example, let K = (6, 10, 15). Here it might take some work, but it shouldn t be hard to prove that K = Z. For example, the fact that 10 and 15 are in K means that 5 = 15 10 is in K. Then knowing that 6 K as well shows that 6 5 = 1 K at which point we have K = Z. We also saw examples of intersections of subgroups. What is (6) (10) (15)? In english, this subgroup would contain all elements of Z which are multiples of 6, 10 and 15. Some thought reveals that such integers are always multiples of 30. That is, (6) (10) (15) = (30). 1.2. C 12. Consider the cyclic group C 12. As a set it looks like C 12 = {e, a, a 2, a 3, a 4, a 5, a 6, a 7, a 8, a 9, a 10, a 11 } where the group operation is a i a j = a i+j subject to knowing a 12 = e. Let s look at all the subgroups generated by single elements. They are (e) = {e} (a) = {e, a, a 2, a 3, a 4, a 5, a 6, a 7, a 8, a 9, a 10, a 11 } (a 2 ) = {e, a 2, a 4, a 6, a 8, a 10 } (a 3 ) = {e, a 3, a 6, a 9 } (a 4 ) = {e, a 4, a 8 } (a 5 ) = {e, a 5, a 10, a 3, a 8, a, a 6, a 11, a 4, a 9, a 2, a 7 } (a 6 ) = {e, a 6 } (a 7 ) = {e, a 7, a 2, a 9, a 4, a 11, a 6, a, a 8, a 3, a 10, a 5 } (a 8 ) = {e, a 8, a 4 } (a 9 ) = {e, a 9, a 6, a 3 } (a 10 ) = {e, a 10, a 8, a 6, a 4, a 2 } (a 11 ) = {e, a 11, a 10, a 9, a 8, a 7, a 6, a 5, a 4, a 3, a 2, a}. Some observations are in order. First of all, notice that the size of each of these subgroups is a divisor of 12. We already knew that would happen since o(c 12 ) = 12 and Lagrange s theorem tells us that the order of a subgroup divides the order of the group. We also see that some cyclic subgroups can be generated by multiple elements. For example, the elements a, a 5, a 7, a 11 all generate the entire group. What are the right cosets of (a 2 )? Since this subgroup has size 6 we expect it to have index 2. Sure enough, we have (a 2 )e = {e, a 2, a 4, a 6, a 8, a 10 } (a 2 )a = {a, a 3, a 5, a 7, a 9, a 11 }.
MATH 370 - EXAMPLES: GROUPS, SUBGROUPS, COSETS 3 Similarly, the subgroup (a 3 ) is expected to have 3 cosets since o(a 3 ) = 4. We have (a 3 ) = {e, a 3, a 6, a 9 } (a 3 )a = {a, a 4, a 7, a 10 } (a 3 )a 2 = {a 2, a 5, a 8, a 11 }. We saw in class that any two right cosets are either disjoint or equal. These three cosets are disjoint, but you should check for example that (a 3 )a = (a 3 )a 7. Again we can look at intersections of subgroups or other subgroups generated by elements. We can just check that (a 2 ) (a 4 ) = (a 4 ) and (a 2, a 5 ) = C 12 since a 5 (a 2 ) 2 = a (a 2, a 5 ). You should verify that in fact all of the subgroups I ve listed above are the only subgroups of C 12. 2. Symmetric Groups 2.1. S 3. We recall that S 3 is generated by two elements, σ and τ where σ 2 = τ 3 = e and τσ = στ 2. As a set we have S 3 = {e, τ, τ 2, σ, στ, στ 2 }. The group S 3 is small enough to list out all of its subgroups. They are H 1 = {e} H 2 = {e, σ} H 3 = {e, στ} H 4 = {e, στ 2 } H 5 = {e, τ, τ 2 } H 6 = {e, τ, τ 2, σ, στ, στ 2 }. Again, these subgroups all have orders which are factors of 6. The group S 3 is not cyclic, in fact it is not even abelian, but it does have the curious property that all of its proper subgroups are cyclic. Knowing all the subgroups of a group makes computing subgroups generated by sets trivial. For example, the subgroup (σ, τ 2 ) must be H 6 since this is the only subgroup of S 3 containing both σ and τ 2. Since H 3 has order 2, we expect H 3 to have three right cosets. Sure enough, we have H 3 e = {e, στ} H 3 σ = {σ, τ 2 } H 3 τ = {τ, στ 2 }. It s worth pointing out that one can form left cosets as well. If you follow through all of the proofs we did in class, you will see that we should also
4 DR. ZACHARY SCHERR expect three left cosets of H 3. As it so happens, the left cosets are different than the right cosets! eh 3 = {e, στ} σh 3 = {σ, τ} τ 2 H 3 = {τ 2, στ 2 }. Sometimes the left cosets and right cosets are all the same. Check that this holds for H 5. These subgroups will be very special, and we will see why in section 2.6 2.2. S 4. We haven t spent too much time working with S 4, but I want to use it to give at least a few examples. We know that S 4 has order 24 and is non-abelian. Let s look at two elements of S 4 and figure out the subgroup they generate. Let r denote the one-to-one correspondence x 1 x 2 x 2 x 3 x 3 x 4 x 4 x 1 and let s denote the one-to-one correspondence x 1 x 2 x 2 x 1 x 3 x 4 x 4 x 3. Then r, s S 4 and we can consider (r, s), the subgroup generated by r and s. You should check that the elements e, r, r 2, r 3, s, sr, sr 2, sr 3 are all distinct elements of S 4 (they represent different functions). Thus these eight elements are all contained in (r, s) but in fact these 8 elements themselves form a group. Thus (r, s) = {e, r, r 2, r 3, s, sr, sr 2, sr 3 }. The easiest way to see that these eight elements are distinct and form a group, other than to do all possible compositions, is to realize that the functions r and s satisfy r 4 = s 2 = e, and rs = sr 1. This looks an awful lot like the relations holding on S 3. You should check that H = (r, s) is a non-cyclic, non-abelian subgroup of S 4 of order 8. Thus we expect it to have index 3.
MATH 370 - EXAMPLES: GROUPS, SUBGROUPS, COSETS 5 Let g be the one-to-one correspondence x 1 x 2 x 2 x 3 x 3 x 1 x 4 x 4. You can check that the three cosets of H in S 4 are H,Hg and Hg 2. To check this, you would just check that these cosets are all disjoint. The group S 4 has many other subgroups, and it might be worthwhile to fool around and see how many you can come up with. 3. More mod. arithmetic Let p be a prime and consider the set J p, J p = {[0], [1],..., [p 1]}. We showed how to defined addition and multiplication on J p and we saw that J p is a group under addition. It is not a group under multiplication since [0] does not have a multiplicative inverse. It is the case, however, that the set J p {[0]} is a group under multiplication. You might not be aware, but this is what you proved in Herstein 1.3.13. You proved that if [a] [0] then [a] has a multiplicative inverse in J p. Let s look at an example. Let p = 13, then the claim is that G = J 13 {[0]} = {[1], [2],..., [12]} is a group. What you may not realize is that this G is actually a cyclic group. We can check this by direct calculation. ([2]) 1 = [2] ([2]) 2 = [4] ([2]) 3 = [8] ([2]) 4 = [3] ([2]) 5 = [6] ([2]) 6 = [12] ([2]) 7 = [11] ([2]) 8 = [9] ([2]) 9 = [5] ([2]) 10 = [10] ([2]) 11 = [7] ([2]) 12 = [1] Thus we see that o([2]) = 12 and G = ([2]). This is great since we can reuse our analysis from studying C 12 earlier! For example, what is the subgroup
6 DR. ZACHARY SCHERR generated by the element [3]? Well we just verified that [3] = ([2]) 4. In our analysis of C 12 we said that the subgroup generated by a 4 was {e, a 4, a 8 }. Replacing a with [2] we see that the subgroup generated by ([3]) is {[1], [3], [9]} = {([2]) 0, ([2]) 4, ([2]) 8 }.