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Home , Semidirect product

Groups of 20

James Pascaleff

November 21, 2019

Here is a solution to the problem of classifying groups of order 20; this was problem 5.1.15 on the homework.

Proposition 1. Let G be a of order 20. Then G is isomorphic to either a semidirect Z5 oα Z4, or a semidirect product Z5 oα (Z2 Z2). × Proof. From the prime factorization 20 22 5, we see that a 2-Sylow has order 4, and a = · 5-Sylow subgroup has order 5. Let P be a 5-Sylow subgroup, and let Q be a 2-Sylow subgroup. Then P Q {e} since the orders of the are relatively prime. We claim that P is normal, which ∩ = is to say that n 1, where n is the number of 5-sylow subgroups. By the third Sylow theorem, n 5 = 5 5 divides 20/5 4, and n 1 (mod 5). The former condition means 1 n 4, and combining this = 5 ≡ ≤ 5 ≤ with the congruence forces n 1. Thus P is normal, and so the product set PQ G is a subgroup. 5 = ⊆ This group is isomorphic to P oα Q, for some α : Q Aut(P) which has order 5 4 20; since G has → × = 20 elements, we must have PQ G. In conclusion, G P oα Q. = =∼ Now, since P has order 5, which is prime, we must have P ∼ Z5, by the classification of groups of 2 = prime order. Since Q has order 2 , either Q Z4 or Q Z2 Z2, by the classification of groups order =∼ =∼ × p2 where p is prime. This completes the proof.

I have instructed the grader to give full marks on this homework problem for a solution that contains something equivalent to Proposition 1. In light of Proposition 1, it remains to understand to look at the two cases (Q Z4 and Q Z2 Z2) =∼ =∼ × and list all of the possible homomorphisms. This takes a bit of work and some knowledge about the autmorphism group of Z5. Proposition 2. Let p be a prime number. Let ϕ : Z Z be an . Then there is some [k] p → p ∈ Z , [k] [0], such that ϕ([x]) [k][x]. In other words, every automorphism is given by multiplication p 6= = by an invertible element [k] Z×. Conversely, every such function is an automorphism. ∈ p Proof. Given ϕ : Z Z , set [k] ϕ([1]). Then for any x with 0 x p, ϕ([x]) ϕ([1] [1] p → p = ≤ < = + + [1]), where there are x terms in the sum. Since ϕ is a homomorphism, this equals [k] [k] [k], ···+ + +···+ where there are x terms in the sum. And that equals [k][x]. Thus ϕ([x]) [k][x] for all [x] Z , and = ∈ p we see that ϕ is completely determined by [k] ϕ([1]). = For the converse direction, observe that the inverse function of [x] [k][x] is [x] [k0][x], where 7→ 7→ [k0] is the multiplicative inverse of [k], i.e., kk0 1 (mod p). ≡ We can take away from this proof a more general fact about homomorphisms whose domain is a cylic group.

Proposition 3. Let G be a group and let ϕ : Z G be a homomorphism. Then ϕ is completely deter- n → mined by ϕ([1]). The element ϕ([1]) G must have order dividing n. ∈ 1 Let us now return to our example. We know Aut(Z5) Z×, which has order 4. The element [1] Z× =∼ 5 ∈ 5 has order 1; the element [2] Z× has order 4; the element [3] Z× has order 4; and the element [4] Z× ∈ 5 ∈ 5 ∈ 5 has order 2. Let us consider the case Q Z4. A homomorphism α : Z4 Z× is completely determined by α([1]), =∼ → 5 which must be an element whose order divides 4. Thus α([1]) may be any of the elements of Z5×. This gives rise to four possible semidirect products, depending on whether α([1]) [1],[2],[3], or [4]. = Let us consider the case Q Z2 Z2. A homomorphism α : Z2 Z2 Z× is determined by ϕ([1],[0]) =∼ × × → 5 and ϕ([0],[1]). These must by elements of order dividing two, so either [1] or [4]. Thus there are again four possible semidirect products, since we have two choices each for ϕ([1],[0]) and ϕ([0],[1]). By this point, we have shown that there are at most eight different groups of order 20. However, we must consider the question of whether this list of eight groups is redundant, which is to say whether some of the groups on the list may be isomorphic to each other. In fact, it is redundant; to see this we can use the result of Exercise 5.4.13. The two semidirect products where α is trivial are abelian, and they are not isomorphic to each other or to any of the semidirect products where α is nontrivial. Let us consider the case Q Z Z . In fact the of Z Z is isomorphic to = 2 × 2 2 × 2 S3, since the elements ([[1],[0]), ([0],[1]), and ([1],[1]) can be permuted arbitrarily. With some more though this shows that the three semidirect products Z5 oα (Z2 Z2) where α is nontrivial are all × isomorphic. Let us consider the case Q Z . There is an automorphism ϕ : Z Z given by ϕ([x]) [ x]. This = 4 4 → 4 = − automorphism plus Exercise 5.4.13 shows that the semidirect product Z5 oα Z4 where α([1]) [2] = is isomorphic to the one where α([1]) [3]. It remains to show that the semidirect product where = α([1]) [2] is not isomorphic to the one where α([1]) [4]. We distinguish them by counting elements = = of order 2. Consider the case Z5 oα Z4 where α([1]) [2]. The product rule in this group is = ([x ],[y ])([x ],[y ]) ([x 2y1 x ],[y y ]) 1 1 2 2 = 1 + 2 1 + 2 Note that the exponentation is computed modulo 5. Look for elements of order 2. These are ([x],[y]) such that ([x],[y])([x],[y]) ([x 2y x],[2y]) ([0],[0]) = + = The possible solutions are ([0],[0]), and ([x],[2]) for any [x] Z . We don’t count the identity, so there ∈ 5 are five elements of order 2. Consider the case Z5 oα Z4 where α([1]) [4]. The product rule in this group is = ([x ],[y ])([x ],[y ]) ([x 4y1 x ],[y y ]) 1 1 2 2 = 1 + 2 1 + 2 Look for elements of order 2. These are ([x],[y]) such that

([x],[y])([x],[y]) ([x 4y x],[2y]) ([0],[0]) = + = The possible solutions are ([0],[0]), and ([0],[2]). Thus there is one element of order 2.

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