Math 4/541 Day 25 1. Previously we showed that given a homomorphism, ϕ, the set of (left) cosets, G/K of the kernel formed a group under the operation akbk = abk. Some observations: We could have just as easily done right cosets because ak and Ka are the same set, namely the set of elements that map to ϕ(a). You can always form the set of (left) cosets of any subgroup. Our Handy Little Coset Lemma was proved for cosets of any subgroup. So we can use it in the general case. The set of cosets of a subgroup, H, always forms a partition of subsets of equal cardinality. Informally, H divides G evenly. Why is this true? First note that for any g in G, gh has the same cardinality as H because for every h in H there is exactly one element of the form gh in gh (by the cancellation law). So all cosets are the same size as H. Second, note that if g is in ah and bh, then g = ah1 and g = bh2 for some h1, h2 in H. This means that ah1 = bh2 which implies that a - 1 b = h1h2-1 is in H. So by HLCL, we know that ah = bh. Thus if two cosets intersect at all, they are the same coset. Finally, it is clear that every element, g, of G is in some coset (namely gh).
2. Theorem (Lagrange): If G is finite then the order of any subgroup H divides G. In fact, G = H [G:H] where [G:H] is the number of cosets of H. Terminology: [G:H] is called the index of H in G. Proof: Already done! We can divide up the elements of G into cosets of H. Each of these cosets has H elements and there are [G:H] of them. 3. More observations: Although we can form left (or right) cosets using any subgroup, they don t always work out as nicely as with the kernel of a homomorphism. Example: Let G = D8 and H = {I, F}. Unlike with a kernel, the left and right cosets of H are not the same. For example RH = {R, FR 3 } but HR = {R, FR}. This is bad because we can t use the notation G/H for both left and right cosets without causing confusion. But it is worse than that. Recall that we defined our operation on cosets of the kernel by akbk = abk. But this operation would not be well defined for cosets of H in this example. Note that RH = FR 3 H = {R, FR 3 } and obviously RH = RH. However our operation gives different answers for RH RH and FR 3 H RH. RH RH = R 2 H = {R 2, FR 2 } but FR 3 H RH = FH = {F, I}. So at least in this case, we cannot form a group of left cosets because our operation doesn t work.
So what property do we need to ensure that this will work? Let s look at our proof that the operation was well defined for cosets of the kernel of a homomorphism: Suppose that ak = ck and bk = dk. We need to prove that akbk = abk is the same as ckdk = cdk. By HLCL, abk = cdk iff (ab) - 1 cd = b - 1 a - 1 cd is in K. However, (also by HLCL) we already know that a - 1 c and b - 1 d are in K. So ϕ(b - 1 a - 1 cd) = ϕ(b - 1 ) ϕ(a - 1 c) ϕ(d) = ϕ(b - 1 )ehϕ(d) = ϕ(b - 1 )ϕ(d) = ϕ(b - 1 d) = eh Thus, (ab) - 1 cd is in K and so abk = cdk. This means our operation is well defined! We wanted to show b - 1 a - 1 cd was in the K and we knew that a - 1 c and b - 1 d were in K. Then we used the definition of the kernel to do away with the a - 1 c in the middle so that we could get b - 1 and d together. But we can t do that in the general case. Let s try to follow our noses to see what property could help us out in the general case. Suppose that ah = ch and bh = dh. If our operation is going to be well defined, we need ahbh = abh to be the same as chdh = cdh. This will be the case if (ab) - 1 cd = b - 1 a - 1 cd is in H. (HLCL) We don t have a homomorphism to plug into now, but we may still be able use the fact that a - 1 c and b - 1 d are in H (HLCL). We know that a - 1 c for some h1 in H. So all have to do is prove that b - 1 h1d is in H. Note that if G were commutative then we would be home free. We could easily show that b - 1 h1d = b - 1 dh1 = h2 h1 for some h2 in H. But really we don t need all of G to be commutative. If everything in H commutes with G, that will also allow us to complete the proof. So we know if will work if H is the center of G (or a subgroup of the center). But we really don t need that much either. It doesn t matter if h1d = dh1. All that matters is that h1d = d times something in H. We could finish the proof like so: b - 1 h1d = b - 1 dh2 = h3 h1 for some h2, h3 in H.
So a sufficient condition for coset multiplication to be well defined is: For any g in G and h in H, hg = gh for some h in H. Definition: A subgroup H is normal in G if for any g in G and h in H, hg = gh for some h in H. I claim that this condition is both necessary and sufficient: Proof: (Sufficient) Suppose H is normal in G. Suppose that ah = ch and bh = dh. By definition, ahbh = abh and chdh = cdh. We need to show these are the same. It is sufficient to show that (ab) - 1 cd = b - 1 a - 1 cd is in H. Well, b - 1 a - 1 cd = b - 1 h1d for some h1 in H. Then by normality, b - 1 h1d = b - 1 dh2 for some h2 in H. And finally since b - 1 d is in H, we have that (ab) - 1 cd is in H as desired. So coset multiplication is well defined. (Necessary) Suppose coset multiplication is well defined. Then for any g in G, and h in H, it is clear that gh = gh and eh=hh were e is the identity of G. Since the operation is well defined, we know that gh = hgh. This implies that g - 1 hg is in H. Which implies that there exists h in H such that g - 1 hg = h which means that hg = gh. So H is normal in G. Normality is the magic condition needed for coset multiplication be well defined. It is (again) easy to show that if H is normal then G/H forms a group under coset multiplication.
Some observations: Since coset multiplication is well defined for the kernel of a homomorphism, we know that the kernel of a homomorphism is always a normal subgroup. You should try to prove this directly using a definition of normality. There are a number of equivalent definitions of normality. You should try to prove that the following are equivalent: o H is normal in G (using the definition above) o For any g in G and h in H, g - 1 hg is in H. o For any g in G, gh = Hg. o For any g in G, g - 1 Hg = H. The second version is probably the most common definition found in textbooks. The third version allows us to use G/H to represent the set of cosets without worrying whether they are left or right cosets. Note that this notation is used even if H is not normal and in those cases you need to pay attention. The fourth version should be familiar from the first exam.