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Cosets, Lagrange's Theorem and Factor Groups

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Cosets, Lagrange's Theorem and Factor Groups 6 Cosets & Factor Groups The course becomes markedly more abstract at this point. Our primary goal is to break apart a group into subsets such that the set of subsets inherits a natural group structure. This sounds strange, but it is precisely what you’ve already encountered when constructing a natural addition on the set of remainders modulo n: in Z6, when we write 3 + 5 = 2, we really mean [3] + [5] = [2], where [3] = f3 + 6n : n 2 Zg is the set of integers whose remainder is 3 when divided by 6. We shall call this set a coset to indicate that it is related to a subgroup of the integers, namely 3Z. That the above addition of sets makes sense and defines a group structure is merely an example of a more general construction. 6.1 Lagrange’s Theorem Before properly defining cosets, we state what you should already have hypothesized, given the huge number of examples you’ve now seen. Cosets will turn out to provide a simple proof. Review any of the subgroup relations we’ve already seen and observe that the order of a subgroup is a divisor of the order of its parent group. This is a general result. Theorem 6.1 (Lagrange). Suppose that G is a finite group. The order of any subgroup H ≤ G divides the order of G. Otherwise said H ≤ G =) jHj jGj We have already proved the special case for subgroups of cyclic groups:1 If G is a cyclic group of order n, then, for every divisor d of n, G has exactly one subgroup of order d. More precisely, if G = hgi has order n, then k ∼ = n • g = Cd where d gcd(n,k) • gk = gl () gcd(n, k) = gcd(n, l) We shall prove the full theorem shortly. For such a simple result, Lagrange’s Theorem is extremely powerful. For instance, you may have observed that there is only one group structure, up to isomor- phism, of prime orders 2, 3, 5 and 7. This is also a general result. Corollary 6.2. There is only one group (up to isomorphism) of each prime order p, namely Cp. Proof. Let p be prime and suppose that G is a group with jGj = p. Since p ≥ 2, we may choose some element x 2 G n feg. Consider the cyclic subgroup generated by x, namely hxi ≤ G. Lagrange =) jhxij divides p =) hxi = 1 or p. Since hxi has at least two elements, its order must therefore be p, from which G = hxi is cyclic. Finally, recall that there is only one cyclic group of each order up to isomorphism, so G = Cp. 1Lagrange’s Theorem is sometimes misremembered as ‘the order of an element divides the order of a group.’ This is really only a statement about the cyclic subgroups of a group G and is not as general as Lagrange-proper. 1 6.2 Cosets and Normal Subgroups The idea for the proof of Lagrange’s Theorem is very simple. Given a subgroup H ≤ G, partition G into several subsets, each with the same cardinality as H. We call these subsets the cosets of H. Definition 6.3. Let H ≤ G and choose g 2 G. The subsets of G defined by gH := fgh : h 2 Hg and Hg := fhg : h 2 Hg are (respectively) the left and right cosets of H containing g. If the left- and right- cosets of H containing g are always equal (8g 2 G, gH = Hg) we say that H is a normal subgroup of G, and write H / G. If G is written additively, the left and right cosets of H containing g are g + H := fg + h : h 2 Hg and H + g := fh + g : h 2 Hg Even though our ultimate purpose is to prove Lagrange’s Theorem (a statement about finite groups), the concept of cosets is perfectly well-defined for any group. Before seeing some examples, we state a straightforward lemma giving some conditions for identifying normal subgroups. Lemma 6.4. 1. Every subgroup of an abelian group G is normal. 2. A subgroup H is normal in G if and only if 8g 2 G, 8h 2 H, ghg−1 2 H. For non-abelian groups, most subgroups are typically not normal (although see example 3 below). Proof. Part 1. should be obvious. For part 2., note first that H / G =) 8g 2 G, h 2 H, gh 2 Hg =) 8g 2 G, h 2 H, ghg−1 2 H Conversely, ghg−1 2 H =) gh 2 Hg =) gH ⊆ Hg. If this holds for all g 2 G and h 2 H, then it certainly applies for g−1, whence g−1hg 2 H =) hg 2 gH =) Hg ⊆ gH We conclude that gH = Hg for all g 2 G and so H is normal in G. Examples 1. Consider the subgroup of Z12 generated by 4. This subgroup is cyclic of order 3: ∼ H = h4i = f0, 4, 8g = C3 Since the group operation is addition, we write cosets additively: for example, the left coset of h4i containing x 2 Z12 is the subset x + h4i = fx + n : n 2 h4ig = fx, x + 4, x + 8g 2 where we might have to reduce x + 4 and x + 8 modulo 12. The distinct left cosets of h4i are as follows: 0 + h4i = f0, 4, 8g = 4 + h4i = 8 + h4i (usually written h4i, dropping the zero) 1 + h4i = f1, 5, 9g = 5 + h4i = 9 + h4i 2 + h4i = f2, 6, 10g = 6 + h4i = 10 + h4i 3 + h4i = f3, 7, 11g = 7 + h4i = 11 + h4i There are only four distinct cosets: notice that each has the same number of elements (three) jZ12j = 12 = Z as the subgroup h4i, and that jh4ij 3 4. The cosets have partitioned 12 into equal-sized subsets. This is crucial for the proof of Lagrange’s Theorem. Observe also that the right cosets of h4i are the same as the left cosets, in accordance with Lemma 6.4). Observe finally that only one of the cosets is a subgroup, the others are merely subsets of Z12. 2. Recall the multiplication table for D3. With a little work, you should be able to compute that the left and right cosets of the subgroup H = fe, m1g are as follows: Left cosets Right cosets eH = m1 H = H = fe, m1g He = Hm1 = H = fe, m1g r1 H = m3 H = fr1, m3g Hr1 = Hm2 = fr1, m2g r2 H = m2 H = fr2, m2g Hr2 = Hm3 = fr1, m3g This time the left and right cosets of H are different (H is not a normal subgroup of D3), but all cosets still have the same cardinality. 1 3. Recall how we proved that the alternating subgroup An ≤ Sn has cardinality jAnj = 2 jSnj. Generalizing this approach, it should be clear that, for any a 2 An and s 2 Sn, we have as even () s even () sa even Otherwise said, for any s 2 Sn, the cosets of An contiaining s are ( An if s even sAn = Ans = Bn if s odd where Bn is the set of odd permutations in Sn. Note that An is a normal subgroup of Sn. 4. The vector space R2 is an Abelian group under addition. The real line R identified with the x-axis, is a subgroup. The cosets of R in R2 are all the sets v + R which are W horizontal lines. More generally, if W is a subspace of a vector space V, then the cosets v + W form sets parallel to W: only the zero coset W = 0 + W is a subspace. In the picture, W is line through the origin in R2; its cosets comprise all the lines in R2 parallel to W (including W itself!). 3 The examples should have suggested the following Theorem. Theorem 6.5. The left cosets of H partition G. Before reading the proof, look at each of the above examples and convince yourself that the result is satisfied in each case. The strategy is to define an equivalence relation, the equivalence classes of which are precisely the left cosets of H. Proof. Define a relation ∼ on G by x ∼ y () x−1y 2 H. We claim that ∼ is an equivalence relation. − Reflexivityx ∼ x since x 1x = e 2 H. X Symmetryx ∼ y =) x−1y 2 H =) (x−1y)−1 2 H, since H is a subgroup. But then −1 y x 2 H =) y ∼ x. X Transitivity If x ∼ y and y ∼ z then x−1y 2 H and y−1z 2 H. But H is closed, whence −1 −1 −1 x z = (x y)(y z) 2 H =) x ∼ z. X The equivalence classes of ∼ therefore partition G. We claim that these are precisely the left cosets of H. Specifically, we claim that x ∼ y () x and y lie in the same left coset of H () xH = yH However, note that x ∼ y () x−1y 2 H () y 2 xH. It follows that yH ⊆ xH. By symmetry, x ∼ y () y−1x 2 H () xH ⊆ yH. We conclude that x ∼ y () xH = yH as required. Aside: Partitions and Subgroups Each of the three conditions that ∼ be an equivalence relation corresponds to one of the properties characterizing H as a subgroup of G: ReflexivityH contains the identity. SymmetryH is closed under inverses. TransitivityH is closed under the group operation. It is precisely the fact that H is a subgroup which guarantees a partition. For example, if H is merely the subset H = f0, 1g of jZ3 = f0, 1, 2g, then its left ‘cosets’ are 0 + f0, 1g = f0, 1g, 1 + f0, 1g = f1, 2g, 2 + f0, 1g = f2, 1g, which do not partition Z3.
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