HW2 Solutions. Let d be the distance from the point P to the origin. Then the orbit of P is the circle of radius d centered at the origin.

Transcription

1 HW2 Solutions Section 16 13) Let G be the additive group of real numbers Let the action of θ G in the real plane R 2 be given by rotating the plane counterclockwise about the origin through θ radians Let P be a point other than the origin in the plane (a) Show that R 2 is a G-set Note that elements of our set are points other than the origin in the plane R 2 (the origin is fixed) To show that R 2 is a G-set we need to show that the action defined above satisfy the two criteria for a group action The additive identity is 0 R G Rotating by 0 radians means that the plane stays fixed In particular every point in the plane is fixed and so 0 P P Now let θ 1 θ 2 G We then have to show that (θ 1 + θ 2 ) P θ 1 (θ 2 P ) Note that acting on P by θ 1 + θ 2 is exactly the same as first rotating by θ 2 and then by θ 1 And so the equation holds (b) Describe geometrically the orbit containing P Let d be the distance from the point P to the origin Then the orbit of P is the circle of radius d centered at the origin (c) Find the group G P The group G P is the subgroup of G R that fixes the point P Note that rotating by integer multiples of 2π rotates the plane back to its original position Therefore G P 2π R Extra Problem: Let G act on itself by conjugation (a) Show that every element in the same conjugacy class has the same order Recall that the order of an element is the smallest positive integer n Z + such that a n e where e G is the identity element Let a G such that a n for

2 some positive integer n Z + Also let b G be in the conjugacy class of a That means that there exists g G such that b gag 1 We then consider b n (gag 1 ) n (gag 1 )(gag 1 ) (gag 1 ) }{{} n of these ga(g 1 g)a(g 1 g) (g 1 g)ag 1 ga n g 1 geg 1 e But we re not done because this just shows that the order of b is n Assume that there exists m Z + with m < n such that b m e Since b gag 1 we can solve for a and get a g 1 bg Applying the exact same argument as above we would get that a m e But this contradicts the assumption that the order of a is n (remember n is the smallest such integer such that a n e) So we indeed have b n (b) Determine the conjugacy classes of the symmetric group on 3 letters I m going to use cycle notation here S 3 {(1) (1 2) (2 3) (1 3) (1 2 3) (1 3 2)} You can use the book to figure out to what ρ s and µ s these elements correspond By part (a) we can cut down our work a little bit since we now know that only elements of the same order might be in the same conjugacy class So at the very least we know that the identity is in its own conjugacy class There are only two elements of order 3 namely (1 2 3) and (1 3 2) So we think these guys might be in the same conjugacy class So we just have to find an element σ S 3 such that σ(1 2 3)σ 1 (1 3 2) Note that all the cycles of length 2 are self-inverse (1 2)(1 3 2)(1 2) 1 (1 2)(1 3 2)(1 2) (1 2)(2 3) (1 2 3) So (1 2 3) and (1 3 2) are indeed in the same conjugacy class Now we wonder if all the order 2 elements are in the same conjugacy class So we just guess and check (2 3)(1 2)(2 3) (2 3)(1 2 3) (1 3) And so we know that (1 3) is in the same conjugacy class as (1 2) Similarly we have that (1 3)(1 2)(1 3) (1 3)(1 3 2) (2 3) and so (2 3) is also in the same conjugacy class as (1 2) as well Therefore there are three conjugacy classes {e} {(1 2) (2 3) (1 2)} {(1 2 3) (1 3 2)}

3 Section 18 20) Consider the matrix ring M 2 (Z 2 (a) Find the order of the ring that is the number of elements in it Note that there are only two elements in Z 2 {0 1} In a 2 2 matrix there are 4 entries and for each entry there are two possiblities (either 0 or 1) And so M 2 (Z 2 ) (b) List all units in the ring Hopefully you recall from matrix algebra that a matrix has an inverse if and only if it has nonzero determinant But you only know this when the entries of your matrix are real numbers It turns out to be true in this case since Z 2 is a field but since we haven t shown that we can t use that fact So we will use brute force We first [ note that the identity element in this ring is indeed the identity matrix I which is its own inverse and therefore a unit 0 1 Suppose that the matrix has a row of 0 s [ [ [ 0 0 c d a b e f 0 0 ac + be ad bf which can never be the identity matrix no matter what a b c d e f are We can make similar arguments for if the second row or either column consists entirely of 0 s Therefore the only things left to check are the elements are [ 0 1 [ [ 1 1 [ and [ 1 1 So let s take one of these guys and multiply it by an arbitrary matrix and set it equal to the identity matrix I Suppose we have that [ [ [ [ 0 1 a b c d c d a + c b + d For this equation to be true we must have that [ c 1 and d [ 0 giving us that a 1 and b 1 Thefore we know that both and are units 1 1 since they are inverses of each other We can make similar arguments for the other 3 matrices in question and will find that they are all self-inverse

4 38) Show that a 2 b 2 (a + b)(a b) for all a and b in a ring R if and only if R is commutative Remember that a ring R being commutative means that the multiplication is commutative Let a b R be arbitrary and let s take the right-hand side of the equation in question and expand it out By the distributive laws we have Furthermore we have that (a + b)(a b) a(a b) + b(a b) a 2 ab + ba b 2 a 2 ab + ba b 2 a 2 b 2 if and only if ab + ba 0 which is true if and only if ba ab Note that this is exactly what it means for R to be commutative since our choice of a and b were arbitrary Since all of the statements above were if and only if statements we are done 44) An element a of a ring R is idempotent if a 2 a (a) Show that the set of all idempotent elements of a commutative ring is closed under multiplication Let a b R be idempotents that is a 2 a b 2 b Then to show that the set of all idempotents is closed under multiplication we just need to show that ab is also idempotent that is (ab) 2 ab Note that (ab) 2 (ab)(ab) a(ba)b Since R is commutative we have that ba ab and so (ab) 2 (ab)(ab) a(ba)b a(ab)b a 2 b 2 ab since a and b are idempotents by assumption (b) Find all idempotents in the ring Z 6 Z 12 Note that an element (a b) Z 6 Z 12 is idempotent if and only if (a b) 2 (a b) Note that (a b) 2 (a b)(a b) (a 2 b 2 ) and so (a b) is idempotent if and only if a 2 a in Z 6 and b 2 b in Z 12 We proceed by just checking the squares of all elements in these rings In Z 6 we have:

5 In Z 12 we have: Therefore the idempotents of the ring Z 6 Z 12 are the sixteen ordered pairs in the set { } { } 48) Show that a subset S of a ring R gives a subring of R if and only if the following hold: i) 0 S; ii) (a b) S for all a b S; iii) ab S for all a b S Note that this is an if and only if statement so we have to show the implication both ways Let us first assume that S is a subring of R Therefore (i) and (iii) automatically hold If b S then we must have b S since S is a subgroup under addition Therefore (a b) a + ( b) S for all a b S verifying criteria (ii) Now let us assume that (i) (ii) and (iii) hold We will then show that S is a subring Note that we have that the associativity of addition and multiplication are inherited from R We now show that S is a subgroup under addition Since 0 S then for all a S we have that 0 a a S that is S is closed under additive inverses Therefore we also have that for all a b S a + b a ( b) S again by criterion (ii) And so we indeed have that S is a subgroup under addition All that s left to show then is that S is closed under mulitplication but that is exacly criterion (iii)