Math 5111 (Algebra 1) Lecture #17 of 24 ∼ November 9th, 2020 Products of Subgroups Products of Subgroups Semidirect Products This material represents x3.4.3-3.4.4 from the course notes. Products of Subgroups, I We proved earlier that every finitely generated abelian group decomposes as a direct product of cyclic groups. This result tells us that finitely generated abelian groups can be built up from subgroups by taking products. We can often piece other groups together from subgroups in a similar way. We would like to study how to do this, because it will give us more ways to construct finite groups and classify groups of a given order. Products of Subgroups, II If H and K are subgroups of G, then we can certainly consider the subgroup hH; Ki generated by H and K. However, the elements in this subgroup are hard to write down in general, since they are words of arbitrary length in the elements of H and K. If elements from H and K commute with one another, then by rearranging the elements in the word and using the fact that H and K are closed under multiplication, we can reduce any word to a product of the form hk for h 2 H and k 2 K. We will now look at the same set of elements for arbitrary subgroups: this is the idea of the product of two subgroups. Products of Subgroups, III Definition If H and K are subgroups of G, then the product HK is the set HK = fhk : h 2 H; k 2 Kg. The product of two subgroups is not necessarily a subgroup of G. For example, for H = f1; (1 2)g, K = f1; (1 3)g in G = S3, the product HK = f1; (1 2); (1 3); (1 3 2)g, which is not a subgroup of G. However, in some cases HK will be a subgroup: for example, with H = f1; (1 2)g and K = f1; (3 4)g in G = S4, then HK = f1; (1 2); (3 4); (1 2)(3 4)g is indeed a subgroup of G. Products of Subgroups, IV We have various properties of subgroup products: Proposition (Products of Subgroups) Let G be a group and H and K be subgroups of G. #H · #K 1. If H and K are finite, then #(HK) = . #(H \ K) 2. The product HK is a subgroup of G if and only if HK = KH. 3. If H ≤ NG (K) or K ≤ NG (H), then HK is a subgroup of G. 4. If H or K is normal in G, then HK is a subgroup of G. 5. If both H and K are normal in G, and H \ K = feg, then HK is isomorphic to the direct product H × K. 6. If np = 1 for every prime p dividing #G, then G is the (internal) direct product of its Sylow subgroups. Such groups are called nilpotent groups. Products of Subgroups, V Proofs: #H · #K 1. If H and K are finite, then #(HK) = . #(H \ K) Observe that HK is a union of left cosets of K: specifically: HK = [h2H hK. Thus we need only count how many distinct left cosets are obtained, since each left coset has cardinality #K. Consider the action of H by left multiplication on the left cosets of K in HK: by definition, there is a single orbit for this action. Products of Subgroups, VI Proofs: #H · #K 1. If H and K are finite, then #(HK) = . #(H \ K) Notice that the stabilizer of the left coset eK is the set of h 2 H with h · eK = eK, which is equivalent to saying h 2 K. Thus, the stabilizer is simply the set of h 2 H such that h 2 K, which is to say, it is the intersection H \ K. So by the orbit-stabilizer theorem, the size of the orbit is equal to the index [H : H \ K]. This means #H · #K #(HK) = #K · [H : H \ K] = , as claimed. #(H \ K) Remark: If H or K is infinite, then trivially HK is also infinite. We also emphasize that HK is not assumed to be a subgroup here. Products of Subgroups, VII Proofs: 2. The product HK is a subgroup of G if and only if HK = KH. First suppose HK = KH. Let g = hk and g 0 = h0k0 be elements of HK, with h; h0 2 H and k; k0 2 K. Then since HK = KH, the element kh0 2 KH is of the form h00k00 for some h00 2 H and k00 2 K. Then gg 0 = hkh0k0 = h(kh0)k0 = h(h00k00)k0 = (hh00)(k00k0) 2 HK. Likewise, g −1 = k−1h−1 2 KH = HK. Since the identity e = ee is clearly in HK, this means HK is a subgroup of G. Products of Subgroups, VIII Proofs: 2. The product HK is a subgroup of G if and only if HK = KH. Conversely, suppose HK is a subgroup. Then since H and K are both in HK, we have hH; Ki = HK and so KH ⊆ hH; Ki = HK. For the other containment, suppose k 2 K and h 2 H. Then we have h−1k−1 2 HK, so since HK is closed under inverses, we see (h−1k−1)−1 = kh must be in HK for any k; h. Thus, HK ⊆ KH, and so in fact HK = KH. Products of Subgroups, IX Proofs: 3. If H ≤ NG (K) or K ≤ NG (H), then HK is a subgroup of G. Suppose H ≤ NG (K), and let h 2 H and k 2 K. By hypothesis, hkh−1 2 K, and therefore we can write hk = (hkh−1)h 2 KH. Thus, hk 2 KH, and so HK ⊆ KH. Likewise, kh = h(h−1kh) 2 HK, and so KH ⊆ HK. We therefore have KH = HK, and so HK is a subgroup of G by (2). The case where K ≤ NG (H) is essentially identical. Products of Subgroups, X Proofs: 4. If H or K is normal in G, then HK is a subgroup of G. If H is normal in G, then NG (H) = G. Thus, trivially K ≤ NG (H). So by (3), HK is a subgroup of G. Likewise, if K is normal in G, then H ≤ G = NG (K), so again by (3), HK is a subgroup of G. Products of Subgroups, XI Proofs: 5. If both H and K are normal in G, and H \ K = feg, then HK is isomorphic to the direct product H × K. Since H is a normal subgroup of G, by (4) that means HK is a subgroup of G. We first show that the elements of H commute with the elements of K. To see this, observe that if h 2 H and k 2 K, then hkh−1k−1 = (hkh−1)k−1 is an element of K, since hkh−1 2 K since K is normal in G. But hkh−1k−1 = h(kh−1k−1) is also an element of H, since kh−1k−1 2 H since H is normal in G. This means hkh−1k−1 2 H \ K, and so hkh−1k−1 = e, meaning that hk = kh: thus, h and k commute. Products of Subgroups, XII Proofs: 5. If both H and K are normal in G, and H \ K = feg, then HK is isomorphic to the direct product H × K. Next, we claim that every element of HK can be written uniquely in the form hk with h 2 H and k 2 K. To see this suppose hk = h0k0 for h; h0 2 H and k; k0 2 K. Then (h0)−1h = k0k−1. But the left-hand side is an element of H while the right-hand side is an element of K, so by the assumption H \ K = feg, this common element must be the identity e. Thus (h0)−1h = e = k0k−1 and so h0 = h and k0 = k, meaning h and k are unique. Products of Subgroups, XIII Proofs: 5. If both H and K are normal in G, and H \ K = feg, then HK is isomorphic to the direct product H × K. Therefore, we have a well-defined map ' : HK ! H × K mapping hk to the ordered pair (h; k). It is a group homomorphism because if g = hk and g 0 = h0k0 then '(gg 0) = '(hkh0k0) = '(hh0kk0) = (hh0; kk0) = '(hk)'(h0k0) = '(g)'(g 0), where we used the fact that h0 and k commute. Finally, ' is trivially injective (since (h; k) = (e; e) implies hk = e) and surjective (by definition of HK) and so it is an isomorphism. Products of Subgroups, XIV A brief interjection about some terminology in this last situation. If both H and K are normal in G, and H \ K = feg, then HK is isomorphic to the direct product H × K. Under these hypotheses, we call the subgroup HK the internal direct product of H and K, and call the group H × K the external direct product of H and K. The difference is irrelevant as a practical matter, but the distinction is that the internal direct product is defined inside a group that already contains H and K as subgroups, whereas the external direct product is an explicit construction of a new group using the Cartesian product. Products of Subgroups, XV Proofs: 6. If np = 1 for every prime p dividing #G, then G is the (internal) direct product of its Sylow subgroups. The intersection of two Sylow subgroups with different primes is trivial by Lagrange's theorem, since the order of their intersection divides the order of each group.