# Algebra SEP Solutions

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1 Algebra SEP Solutions 17 July (January 2017 problem 1) For example: (a) G = Z/4Z, N = Z/2Z. More generally, G = Z/p n Z, N = Z/pZ, p any prime number, n 2. Also G = Z, N = nz for any n 2, since Z/nZ has non-identity elements of finite order and Z does not. Also the free groups F n, by a similar argument. (b) Any nonabelian group of prime power order, such as the quaternion group Q 8. (c) Since the commutator subgroup is normal, any nonabelian simple group, such as A n for n 5. Also the trivial group. (d) The dihedral group D p of order 2p, where p is any odd prime. (e) S 3 acting on X = S 3 (which has order 6) by left translation. 2. (January 2013 Problem 1) A finite group G is said to have property C if, whenever g G and n is an integer relatively prime to the order of G, g and g n are conjugate in G. (a) Give infinitely many non-isomorphic finite groups which have property C. We claim that the symmetric groups S m satisfy property C. Let n be relatively prime to m!. We will first show that for any r m, and any r-cycle g, g m is also an r-cycle. Assume without loss of generality that g = (12... r). Then g m is the permutation which takes 1 to 1 + m modulo r, takes 1 + m modulo r to 1 + 2m modulo r, and so on. It remains to show that 1 + im is distinct modulo r for 0 i r 1. Indeed, if 1 + im = 1 + jm (mod r), then r (i j)m, which is impossible as (i j) < r and m is relatively prime to r. We recall that two elements of S n are conjugate to each other if and only if they have the same cycle structure, and thus the cycles satisfy the hypotheses of property C. To finish the proof we observe that any permutation can be written as a product of cycles which commute with each other. Raising such a permutation to a power relatively prime to m!, and applying what we have already shown, we would conclude that the resulting permutation had the same cycle structure. This concludes the proof. (b) Give infinitely many non-isomorphic finite groups which do not have property C. One way to think about this is to realize that conjugation is trivial in an abelian group. Let G n = x be the cyclic group on n > 2 elements, and let 1 < m < n be any integer relatively prime to n. Then x m x, showing that G n does not satisfy C. (c) Show that if G has property C and ρ : G GL n (C) is a homomorphism, then the trace of ρ(g) is in Q for all g G. We break the solution into multiple steps. 1

2 Step 1. If G is any finite group, then the eigenvalues of ρ(g) are G -roots of unity. Because G is a finite group, we know that g G = 1. Thus, 1 = ρ(1) = ρ(g G ) = ρ(g) G. The eigenvalues of ρ(g) G are precisely the G -powers of the eigenvalues of ρ(g), and thus the claim is proven. Because the trace of a matrix is the sum of its eigenvalues, it follows that the trace of ρ(g) lands in the field L := Q(ζ), where ζ is a primitive G -root of unity. Step 2. The Galois group Gal(L Q) consists of the automorphisms ζ ζ m, where m is relatively prime to G. Note that any automorphism of L must send ζ to some other primitive root of unity. It is clear that the other primitive roots of unity are of the form ζ m, where m is relatively prime to G, as the generators of any cyclic group are such powers of any fixed generator. It therefore remains to show that any assignment of the form ζ ζ m defines an automorphism. By the automorphism extension theorem it suffices to show that all primitive G -roots of unity are roots of the same irreducible polynomial over Q. This is a well known fact from field theory. Step 3. We are now ready to put everything together and finish the problem. Let g G, and let m be relatively prime to G. By assumption g m is conjugate to g, and therefore ρ(g) and ρ(g) m have the same trace. By what we have shown we may write tr(ρ(g)) = j for some i ζij j. Letting σ Gal(L Q) be the automorphism ζ ζ m, we find that tr(ρ(g)) = tr(ρ(g) m ) = (ζ m ) ij = σ( ζ ij ) = σ(tr(ρ(g))). j j Thus, tr(ρ(g)) is fixed by all elements of Gal(L Q), whence it is rational. 3. (August 2014 Problem 3) Let G be a finite group. (a) If H is a proper subgroup of G, show that there is some element x G which is not contained in any subgroup conjugate to H. Let X be the set of subgroup conjugates of H. Then G acts on X via conjugation, and this action has one orbit. The orbit-stabilizer theorem then tells us that X = [G : N G (H)], where N G (H) is the normalizer of H in G. We note that every conjugate of H has at least the identity element in common, and therefore summing up the sizes of each conjugate individually certainly over counts the total number of elements spanned by the conjugates. In particular, if we assume that these conjugates span all of G then, G < H G / N G (H), so H > N G (H) Because H N G (H), we obtain a contradiction. (b) A maximal subgroup of G is a proper subgroup which is not contained in any other proper subgroup. Derive from the first part of the problem that if all maximal subgroups of G are conjugate, G must be cyclic. Let x G be an element which is not contained in any maximal subgroup. Such an element exists because of the previous part. Set H = x and assume that H G. Let X be the set of proper subgroups of G containing H, equipped with the subgroup order, and observe that X. Because G is of finite order, any chain in X must have finite length, and therefore has an upper bound. By Zorn s lemma we conclude that X must have a maximal element, which is clearly a maximal subgroup of G. This is 2

3 a contradiction. 4. (January 1991 Problem 5) Let G be a possibly infinite, non-trivial group whose subgroups are linearly ordered by inclusion. In other words, if H and K are subgroups of G, then either H K or K H. (a) Prove that G is an abelian group, and that the orders of the elements of G are all powers of the same prime p. If x, y G then x y (or visa versa), and therefore x is a power of y. In particular, x and y commute. Write o(x) for the order of x. If x has finite order, and p, q o(x) are distinct prime divisors, then clearly x o(x)/p and x o(x)/q are two incomparable subgroups. This shows that every element has prime power order. On the other hand, if o(x) = p n and o(y) = q m, then y has a subgroup of order q showing it can t be contained in x unless p = q. Swapping y for x forces us to conclude that p = q. In the case where o(x) =, the subgroups generated by x 2 and x 3 are clearly incomparable. (b) If G n = {g G g pn = 1}, show that G n p n. Note that the largest possible order of an element of G n is p n. Let g G n have maximal order, and let g be any element of G n. Then either g g or visa versa. Because g was chosen to have maximal order, the latter case must be true. This shows that G n = g has size at most p n. 5. (January 1991 Problem 1) Let G be a finite group having exactly n Sylow p-subgroups for some prime p. Show that there exists a subgroup H of the symmetric group S n such that H also has exactly n Sylow p-subgroups. We begin by recalling the generalized Cayley theorem. If G is a finite group, and H is a subgroup of index n, then there exists a homomorphism φ : G S n whose kernel is contained inside H. Indeed, this is just the map representing the permutation action of G on the set of cosets of H. Let P be any Sylow p-subgroup of G, and set H = N G (P ). Then the Sylow theorems tell us that n = [G : N G (P )], and Cayley s theorem gives us a map G S n, whose kernel is contained inside H. Let G = G/ ker(φ) denote the image of this map. Then by simply looking at cardinalities, it follows that (P ker(φ)) is a Sylow p-subgroup of G. We also know that N G (P ker(φ)) = N G ((P ker(φ)) ) by the correspondence theorem. It follows that [G : N G (P ker(φ))] = [G : N G ((P ker(φ)) )] It therefore remains to show that [G : N G (P ker(φ))] = n. N G (P ker(φ)) = N G (P ). In fact, we will show that We immediately find that if x N G (P ), then xp ker(φ)x 1 = (xp x 1 )(x ker(φ)x 1 ) = xp x 1 ker(φ) = P ker(φ). 3

4 This shows the reverse inclusion. On the other hand the Frattini trick immediately implies that N G (P ker(φ)) = N NG (P ker(φ))(p ) P ker(φ), which is a subgroup of N G (P ). 6. (August 2015 Problem 1) Let groups(n) denote the number of groups of order n up to isomorphism. (a) Let p be a prime number. Show that every group of order p 2 is abelian, and determine groups(p 2 ). (b) Find groups(50) (hint: show that every group of order 50 must have a normal Sylow 5-subgroup. (a) Let G be a group of order p 2. Then G is a p-group, and so its center Z(G) is nontrivial. If Z(G) = G we are done, so assume instead that Z(G) = p. Because the center is a normal subgroup we can consider the quotient G/Z(G). This group has order p and is therefore cyclic. To finish the first part of the problem, we make the following general claim: If G is a group such that G/Z(G) is cyclic, then G is abelian. Let a, b G, and assume that G/Z(G) is generated by the image of an element x G. Then we may write x k a Z(G) and x l b Z(G) for some integers l, k. Therefore, ab = x k (x k a)x l (x l b) = x k l (x l b)(x k a) = x k l x k (x l b)a = ba, where we have repeatedly used the fact that x k a and x l b are in the center. For the second part of the problem, we simply apply the structure theorem for finitely generated abelian groups. It follows that G = Z/p 2 Z or G = Z/pZ Z/pZ. (b) Let G be a group of order 50 = 25 2, and let P be a Sylow 5-subgroup of G. By assumption P has index 2 in G, and must therefore be normal. This implies that P is the unique Sylow 5-subgroup of G. Cauchy s theorem (or the Sylow theorems again) imply that we may find an element x G of order 2. Because x / P, we know that P x = P < x > = 50 = G, and therefore G = P x. In other words, G = P Z/2Z Now we recall the internal construction of the semi-direct product discussed above is equivalent to the following external version. We can think of G as being the product P Z/2Z, paired with a homomorphism φ : Z/2Z Aut(P ), with operation (a, b)(c, d) = (aφ b (c), bd) Counting the number of groups therefore becomes the question of how many homomorphisms φ : Z/2Z Aut(P ) there are. By the previous question we know that P = Z/25Z or P = Z/5Z Z/5Z. The automorphisms of Z/25Z are uniquely determined by a choice 1 n where n is relatively prime to 5. For such an automorphism to have order at most 2, it would have to be the case that n 2 = 1 (mod 25). In particular, we obtain two homomorphisms φ : Z/2Z Aut(Z/25Z), φ(1) = (1 1), φ(1) = (1 1). Note that in the first case, G is just the sum of P with Z/2Z, which is necessarily abelian, while in the second case the group is not abelian (in fact, this is the group D 25 ). This shows that the two groups induced by these automorphisms are distinct. Next assume that P = Z/5Z Z/5Z. Then Aut(P ) = GL(2, 5), and any homomorphism φ : Z/2Z Aut(P ) is therefore uniquely determined by a choice of an element of GL(2, 5) with order at most 2. Any matrix of order at most 2 in GL(2, 5) will have as its eigenvalues 1 and -1, the square roots of unity. The Jordan structure theorem 4

5 implies that any such matrix can be conjugated into Jordan form within GL(2, 5), as its eigenvalues are in the base field Z/5Z. If we compose φ with the automorphism induced by such a conjugation, then we obtain a new homomorphism φ : Z/2Z GL(2, 5). By how it was constructed, it isn t hard to see that the group given by φ will be isomorphic to that given by φ. To conclude, it suffices to consider those homomorphisms φ : Z/2Z GL(2, 5) which send the generator of the prior group to an element of order at most 2 in the latter group, which is in Jordan form. There are three such choices φ(1) = ( ) ( ) , φ(1) =, φ(1) = ( ) ( ) 1 0 Note that the assignment φ(1) = is conjugate to the last of the three choices 0 1 above, and is therefore redundant. If we prove that these three homomorphisms yield distinct choices for G, then we would have finished the problem and shown that group(50) = 5. The first case is the case wherein G is the direct product of P with Z/2Z. Looking at the second group, we see that if a P, then (a, 1) is of order 2. In particular, this choice of homomorphism yields a group which has 25 elements of order 2. On the other hand, the third homomorphism does not have this many elements of order 2. This shows all three induced groups are distinct, as desired. 5

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