Name: Id #: Math 56 (11) - Midterm Test Spring Quarter 016 Friday May 13, 016-08:30 am to 09:0 am Instructions: Prob. Points Score possible 1 10 3 18 TOTAL 50 Read each problem carefully. Write legibly. Show all your work on these sheets. Feel free to use the opposite side. This exam has 6 pages, and 3 problems. Please make sure that all pages are included. You may not use books, notes, calculators, etc. Cite theorems from class or from the texts as appropriate. Proofs should be presented clearly (in the style used in lectures) and explained using complete English sentences. Good luck!
Math 56 (11) - Midterm Test Spring Quarter 016 Page of 6 Question 1. (Total of points) a) (8 points) State the definitions/statements of: A composition series of a group G; A solvable group G. The Jordan-Hölder theorem. b) (6 points) Let G be a group and N G a proper normal subgroup. Supposing G admits a composition series, show there exists a composition series containing N. c) (8 points) Let G be a group and N G. Prove that if G is solvable, then N and G/N are both solvable. Solution. a) The definitions/statements are as follows. A subnormal series (H i ) n i=0 of a group G is a composition series if all the factor groups H i+1 /H i are simple. A group G is solvable if it admits a composition series (H i ) n i=0 for which all the factors H i+1 /H i are abelian. Any two composition (or principal) series of a group G are isomorphic. b) By the Schreier refinement theorem, any composition series for G must be isomorphic to some refinement of the series {e} N G (recall, a composition series admits no proper refinement). Since a subnormal series which is isomorphic to a composition series must itself be a composition series, it follows that there exists a refinement of {e} N G which is a composition series. c) We may assume without loss of generality that N is a proper normal subgroup. Supposing G is solvable, by part b) there exists a composition series which contains N, say {e} = H 0 H 1 H H d = N H d+1 H n = G. Since G is solvable, by the Jordan-Hölder theorem the factors H i+1 /H i of this series must be abelian. It follows that {H i } d i=0 is a composition series for N with abelian factors and so N is solvable. On the other hand, by the correpsondence theorem for normal subgroups we have {e G/N } = N/N H d+1 /N H n /N = G/N so defining K i := H d+i /N for i = 0,..., n d it follows that (K i ) n d i=1 is a normal series for G/N. Finally, note that by the Third Isomorphism Theorem we have K i+1 K i = H d+i+1/n H d+i /N = H d+i+1 H d+i
Math 56 (11) - Midterm Test Spring Quarter 016 Page 3 of 6 and so each K i+1 /K i must be simple and abelian for i = 0, 1,..., n d 1. Thus (K i ) n d i=1 is a composition series for G/N and G/N is a solvable group.
Math 56 (11) - Midterm Test Spring Quarter 016 Page of 6 Question. (Total of 10 points) Prove that no group of order 56 is simple. Solution. a) Suppose G is a group of order 56 = 3 7. Then by Sylow III the number of Sylow 7-subgroups is congruent to 1 mod 7 and divides 56 so there must be either 1 or 8 such subgroups. If there is only one such subgroup, then it must be normal (since for each p the set of Sylow p-subgroups is closed under conjugation) and so we assume there are 8 such subgroups. Since they are cylic, they can only intersect at the identity, and so we see there are 8 6 + 1 = 8 distinct elements of G of order 7. On the other hand, the number of Sylow -subgroups is odd and divides 56, so there must either be 1 or 7 such subgroups. Assume there are 7 of these subgroups, each of which has cardinality 8. Choose two distinct Sylow -subgroups H 1, H. Then H 1 H by Lagrange s theorem, so that H 1 H 16 = 1. But now we have 8+1 = 60 elements of the group, which is a contradiction. Hence there must only one Sylow -subgroup, which is normal.
Math 56 (11) - Midterm Test Spring Quarter 016 Page 5 of 6 Question 3. (Total of 18 points) Let denote the positive fourth root of and consider the extension Q(, i) : Q. a) ( points) Find [Q(, i) : Q]. b) (3 points) Find all the conjugates of and i over Q. c) ( points) Let σ, τ Aut Q(, i) : Q be defined by σ : τ : i σ : i i τ : i i. Write down the elements of the subgroup of Aut Q(, i) : Q generated by σ and τ (e.g. ι, σ, σ,... ), describing each by its action on and i. d) (3 points) To which familiar group is Aut Q(, i) : Q isomorphic? e) ( points) Find the fixed fields of the subgroup {1, σ, τ, σ τ}. Solution. a) Clearly i / Q( ) and is a root of x + 1; consequently, [Q(, i) : Q( )] =. On the other hand, by Eisenstein we see that irr(, Q)(x) = x and so [Q( ) : Q] =. By the Tower Law, it follows that [Q(, i) : Q]. b) The conjugates of over Q are given by {, i,, i }. The conjugates of i over Q are {i, i}. c) The automorphisms belonging to the subgroup generated by σ and τ are: ι ι: σ σ : i σ σ : σ 3 σ 3 : i τ τ : στ στ : i σ τ σ τ : σ 3 τ σ 3 τ : i ι: i i σ : i i σ : i i σ 3 : i i τ : i i στ : i i σ τ : i i σ 3 τ : i i Note that all these automorphisms are distinct and they in fact are all the elements of Aut Q(, i) : Q: since each element must be mapped to a conjugate over Q by any ψ Aut Q(, i) : Q there are at most 8 choices of ψ. d) Aut Q(, i) : Q = D, the dihedral group of symmetries of the square: τ corresponds to reflection across the vertical axis and σ to a π/ anti-clockwise rotation. e) Any element x Q(, i) can be expressed as x = a 1 + a + a 3 + a ( ) 3 + a 5 i + a 6 i + a 7 i + a 8 i( ) 3
Math 56 (11) - Midterm Test Spring Quarter 016 Page 6 of 6 for some a 1,..., a 8 Q. Now we see that σ (x) = a 1 a + a 3 a ( ) 3 + a 5 i a 6 i + a 7 i a 8 i( ) 3 ; τ(x) = a 1 + a + a 3 + a ( ) 3 a 5 i a 6 i a 7 i a 8 i( ) 3 ; σ τ(x) = a 1 a + a 3 a ( ) 3 a 5 i + a 6 i a 7 i + a 8 i( ) 3. Consequently, the fixed field is Q( ).