Basic Definitions: Group, subgroup, order of a group, order of an element, Abelian, center, centralizer, identity, inverse, closed.

Transcription

1 Math 546 Review Exam 2 NOTE: An (*) at the end of a line indicates that you will not be asked for the proof of that specific item on the exam But you should still understand the idea and be able to apply the concept Groups and Subgroups the basics Basic Definitions: Group, subgroup, order of a group, order of an element, Abelian, center, centralizer, identity, inverse, closed If a and b are elements of a group G, then ( ab)!1 = b!1 a!1 For any a in G, the function! a :G " G defined by! a ( x) = ax is a bijection For a finite group G, the multiplication table of G is a Latin Square (*) Theorem For any a in a group G and integers r and s, (*) (i) a r a s = a r + s (ii) a r ( ) s = a rs If x in the group G has order k and a m = e, then k divides m Hint: Let m = kq + r, 0! r < d If G is a group of order n and a is an element of G, then a n = e If a and b are arbitrary elements of a group, then ab and ba have the same order If A and B are subgroups of the group G, then so is A! B In fact if H i is a subgroup of G for each i in some set I, then! H i is a subgroup of G (ie, the intersection of any collection of subgroups is a subgroup) Let a be a particular element of the group G, then C a = {x!g :ax = xa} (the centralizer of a in G) is a subgroup of G (ie, the centralizer of a is the set of all elements of G that commute with a) The set Z(G) = {x!g :yx = xy} (the center of G) is a subgroup of G Note that Z(G) =! a!g C a Let H be a subset of G, then H is a subgroup of G iff H is closed, the identity of G belongs to H and for each x in H, x!1 "H (*) If H is a finite subset of a group G, then H is a subgroup iff it is closed i!i

2 Exercises 1 If A and B are subgroups of the abelian group G with A = n, B = m where n and m are relatively prime integers, then show that AB = nm [ Recall that AB = { ab : a!a,!b!b}] Hint: It is enough to show that the function f :A! B " AB, defined by f (a,b) = ab is a bijection Solution: It is enough to show that the function f :A! B " AB, defined by f (a,b) = ab is a bijection So, suppose that f (a 1,b 1 ) = f (a 2,b 2 ) We must show that (a 1,b 1 ) = (a 2,b 2 )! a 1 = a 2 and b 1 But since, f (a 1,b 1 ) = f (a 2,b 2 ), then a 1 b 1 = a 2 b 2! a 1 a "1 2 b "1 1 But that means that the element x = a 1 a!1 2 b!1 1 belongs both to A and to B Thus the order of x must divide both n and m But that means that x must have order 1, in other words that x is the identity Hence a 1 a!1 2 = e b!1 1 " a 1 = a 2 and b 1 2 In Q 8, let A = {i,! j},!!b = {!1,!i} what is AB Solution: AB = {!i,!! j,!!1,!!k} 3 If G is a group and x 2 = e for every x in G, then G is abelian Solution: Let a and b be any elements of G Then (ab) 2 = e! abab = e! a(abab) = ae! bab = a! b(bab) = ba! ab = ba 4 Suppose that G contains exactly one element a of order 2 Show that a belongs to the center of G Hint: for any x in G consider xax!1 group that has exactly one element of order 2 Solution: xax!1 ( ) 2 Give an example of a ( ) 2 = ( xax!1 )( xax!1 ) = xa(x!1 x)ax!1 = xa 2 x!1 = xx!1 = e Hence either xax!1 = e " a = e, which is impossible since a has order 2, or xax!1 has order 2 in which case xax!1 = a " xa = ax Thus a commutes with x and since x was arbitrary, that means that a commutes with every element of G Hence a belongs to the center of G 5 If G is an abelian group, then H = {x :x 2 = e} is a subgroup of G Solution: Clearly, since e 2 = e, e belongs to H Suppose that a and b are elements of G Then (ab) 2 = abab = aabb = ee = e Hence ab!h, so H is closed Finally, suppose that x is any element of H, then x!1 x!1 "H ( ) 2 = x!2 = ( x 2 )!1 = e!1 = e and so

3 6 For real numbers a, b let T a,b (x) = ax + b { } is a group under composition of functions F = T a,b : a,b are real numbers Solution: This follows from the parts below T a,b!t c,d = T, Solution: T a,b!t c,d (x) = T a,b ( T c,d (x)) = T a,b (cx + d) = a(cx + d) + b = acx + ad + b = T ac,ad + b (x) And so, T a,b!t c,d = T ac,ad +b which is in F and so F is closed What is the identity of F Solution: From the lines above, T a,b!t 1,0 = T a,b = T 1,0! T a,b, and so the identity is T 1,0 ( T a,b )!1 = T 1 a,! b a Solution: (Just try it) H = { T a,b : a,b are real numbers, a is rational } is a subgroup of F Cyclic Groups Basic Definitions: generator, cyclic group, G = < g > Every cyclic group is commutative If G and H are any cyclic groups of order n, then G is isomorphic to H (*) Consequently, every cyclic group on n elements is isomorphic to Z n (*) The only infinite cyclic group (up to isomorphism) is the integers under addition (*) Every subgroup of a cyclic group is cyclic If k!z n is relatively prime to n, then k generates (Z n, +) (*) For a cyclic group G = g of order n, G = {e,!e 2,!,,!e n!1 } (*) LaGrange s Theorem and Its Consequences Theorem (LaGrange) Let G be a finite group of order n and H a subgroup of order k, then k divides n (*)

4 So, G = H [ G:H ], where [ G:H ] = G H is the index of H in G (*) For any subgroup H of the group G, the relation x! y mod H " x #1 y $H is an equivalence relation The equivalence class of any element x of G is [x] = xh the left coset of x in G The cardinality of any left coset xh is the same as that of H If x is an element of the finite group G then the order of x divides the order of G If G is a group of prime order, then G is cyclic Theorem (Euler) If a and n are positive, relatively prime integers, then a!( n ) " 1 mod n Theorem (Fermat) If n, p!z +, p is a prime and / p / n, then a p!1 "1 mod p (Note: / p / n means that p does not divide n) Exercises 1 Let A and B be subgroups of G with A = n, B = m where n and m are relatively prime integers Then show that A! B = {e} Solution: suppose that a belongs to both A and B Then the order of a must divide both n and m and hence the order of a is 1 and so a is the identity 2 Show that if G is an abelian group of order 2n, and n is odd, then G has exactly one element of order 2 (This uses LaGrange s Theorem) Hint: We know from a previous exercise that G must contain at least one element of order 2 You need to show here that there cannot be another one Solution: Suppose that G has (at least) two distinct elements a and b of order 2 We will show that this produces a contradiction Then it is easy to check that H = {e, a, b, ab} is a subgroup of G (Recall that if H is a finite subset of a group, then it is a subgroup iff it is closed) Hence, by LaGrange s Theorem, H divides G Hence 4 divides 2n which is impossible since n is odd 3 (a) List all the left cosets of {-1, 1} in the Quaternions (b) What is the order of i in the Quaternions Solution: 4

5 4 Suppose that G is a finite group and K is a subgroup of H and H is a subgroup of G Then show that [ G : K ] = [ G : H ][ H : K ] Solution: See problem set 15 Permutations, Permutation Groups and Cayley s Theorem Basic Definitions: permutation, cycle, transposition Be able to express a permutation as a product of disjoint cycles Be able to find powers and the inverse of a given permutation If!, " are disjoint cycles, then! " = "! (*) Every permutation is the product of disjoint cycles and this representation is unique up to the order of the factors (*) If a permutation is written as a product of disjoint cycles, then its order is the the least common multiple of the lengths of the cycles Example: The order of ( 1,!5,!4)(7,!3,!2,!8) is 12, the order of ( 3,!4) 5,!7,!1 and the order of 3,!5 Exercises ( )( 2,!1,!4) ( 6,!8,!7) ( 9,!10) is 6 1 Find a permutation! such that!(1, 2, 3, 4) = (3,1, 2, 4) Solution:! = (3,1,2,4) 4,3,2,1 ( ) = ( 1,3,4 ) 2 Find a permutation! "S 8 that is not a cycle and such that! 3 = (1, 2, 3, 4) 3 Solution: We may choose! = (1, 2, 3, 4) 5, 6, 7 ( ) 3 (a) What is the order of! = ( 3,!1,!6,!5,!7) "S 8 ( ) is 6, " (b) Express! = % # $ & ' as a product of disjoint cycles (c) What is! 2!!! 3 " (c) Let! = % # $ & ' What is the product!" 4 How many distinct r-cycles with r < n are there in S n Hint: The number of distinct ways to arrange r objects in a circle is (r!1)!! n Solution: # $! & ( r '1)! So, in particular, there are # n$ & transpositions " r% " 2%

6 5 Let a be an integer in {1, 2, 3,, n}, then show that H a = {! :!(a) = a} is a subgroup of S n Solution: Just do it 6 (a) Show that every cycle is a product of transpositions Hint: ( 1,!2,!3,!4) = ( 1,!4) ( 1,!3) ( 1,!2) (b) Show that every permutation is a product of transpositions Solution: A simple consequence of the the fact hat every permutation is a product of cycles 7 What is the index of 1,!2,!3 ( )( 4,!5) in S 8 Solution: 8! 7! 6! 5! 4 = Show that if! "S 5, then! 120 = i (i is the identity permutation) Note that this was originally posted as! "S 9 Solution: The order of S 5 is 120 Can you show that there is a! "S 9 for which the conclusion in (8) fails 9 (a) Does there exist an element of order 15 in S 12 Solution: Yes, for example (1,!2,!3)(4,!5,!6,!7,!8) (b) Does there exist an element of order 13 in S 12 Solution: No If there were such an element then 13 would have to divide 12! 10 (Harder) What is the centralizer of ( 1,!2) in S 4 Solution: You can verify that C [(1,!2)] = {! "S 4 :!(1) = 1 and!(2) = 2!or!!(1) = 2 and!(2) = 1 } Isomorphisms 1 Show that f :( R, + )! C *, " Hint: use the relations cos(x + y) = cos x cos y! sin x sin y, sin(x + y) = sin x cos y + sin y cos x Solution: The hint tells it all ( ) defined by f (x) = cos x + isin x is an isomorphism 2 Let G be a group of order n and m an integer relatively prime to n Show that the function f :G! G, defined by f (x) = x m is a bijection Determine a condition on G that would make f a homomorphism as well Hint: Show that f is onto then it will also be 1-1 since G is finite Solution: Let a and b be integers such that an + bm = 1 Then for any y in G, y = y an +bm = (y n ) a (y b ) m = (y b ) m = f (y b ) and so since y was arbitrarily chosen in G, f is onto

7 The Euler phi-function!(p) = p " 1, and!(p n ) = (p " 1)p n"1 where p is a prime(*)!(nm) =!(n)!(m) if n and m are relatively prime (*) If p is a prime and a and b are integers and p divides ab, then p divides a or p divides b(*) If n and m are relatively prime, then there exist integers a and b such that an + bm = 1 For any positive integers n and d there exists integers q and 0! r < d, such that n = dq + r(*)