# Solutions Cluster A: Getting a feel for groups

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1 Solutions Cluster A: Getting a feel for groups 1. Some basics (a) Show that the empty set does not admit a group structure. By definition, a group must contain at least one element the identity element. (b) Show that the identity element of a group G is unique. (That is, if two elements 1 and 1 satisfy the defining property of the identity element, then 1 = 1.) If 1 is the identity, then it must satisfy the equation 1g = g1 =g for all g. In particular, if 1 = g, we must have 11 =1. On the other hand, if 1 is also the identity element, we must have 11 = 1. By transitivity, we conclude1 = 1. (c) Given an element g G, show that g 1 is unique. (That is, given elements h, h satisfying the defining property of g 1, show that h = h.) Suppose h and h are both inverses to g. Then gh = 1. By multiplying both sides of the equation by h on the left, we obtain h(gh )=h. But by associativity, the lefthand side becomes h(gh )= (hg)h =1h = h. By transitivity of equality, we have that h = h. 3

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3 Homework 2 1. Group homomorphisms versus maps of sets Let φ 3 : Z Z be the map φ 3 (n) =3n. So for instance, φ(2) = 6. We think of the integers as a group under addition. (a) Show that φ 3 is a group homomorphism. By distributivity of multiplication over addition, k(a + b) = ka + kb for all k, a, b Z. Hence φ 3 (a + b) = 3(a + b) =3a +3b = φ 3 (a)+φ 3 (b). (b) Show that there exists a map of sets ψ : Z Z such that ψ φ 3 =id Z. There are many such choices. For instance, by the division algorithm, we can uniquely write any number a Z as a =3k + r where k is an integer and r is some integer between 0 and 2. Setting ψ(a) =k clearly satisfies ψ(3k) = k. (c) Show that no choice of such a ψ can be a group homomorphism. In general, note that = n, where the addition occurs n times. Hence any group homomorphism ψ : Z Z must satisfy ψ(1) ψ(1) = ψ(n), where the addition again occurs n times on the lefthand side. But this means that ψ(n) mustbedivisiblebyn. In contrast, if ψ φ 3 (1) = 1, we must have that ψ(3) = 1, and 1 is not divisible by 3. (d) For any integer k, define a map of sets φ k : Z Z by φ(n) =kn. Show this defines a group homomorphism from Z to Z. Determine all k for which this map is an isomorphism. We showed that this is a homomorphism in general in part (a) above. By our solution to part (c), we see that any homomorphism ψ must send φ k (1) = k to a number divisible by k. On the other hand, ψ φ k = id means ψ(φ k (1)) = 1. The only numbers dividing 1 are k = ±1, so we conclude k must be equal to ±1. In either case ψ k is a bijection, as the function n n and n n are both bijections. 5

4 Fall 2014 Math 122 Homework 2. The sign representation (a) Let S n be the symmetric group on n elements. (Automorphisms of a set of n elements.) For every element σ S n,letφ(σ) bethen n matrix which sends the standard basis vector e i R n to the vector e σ(i).show that the assignment φ : S n GL n (R) is a group homomorphism. Let σ, σ S n. By definition, φ(σ) sendse i to e σ(i), and likewise for φ(σ ). Recall that if two matrices represent two linear transformations T and T, then the product of the two matrices represent the composition of the linear transformations T and T. Hence the product matrix φ(σ )φ(σ) sendse i to φ(σ )(φ(σ)(e i )) = φ(σ )(e (σ(i)) )=e σ (σ(i)) = e (σ σ)(i). A linear transformation (and hence its matrix) is determined by what it does on a set of basis vectors, and we ve seen that φ(σ )φ(σ) represents the linear transformation sending e i to e σ σ(i). By definition, this is the linear transformation represented by φ(σ σ). Hence we have a group homomorphism. (b) List every element σ S 3 and write out the matrix φ(σ) for each of them. Now that we have cycle notation, we can list the elements as follows: id, (12), (23), (13), (123) (132). In corresponding order, the matrices φ(σ) are given by , 1 0 0, 0 0 1, , Note that we see that the ith column of the matrix is given by the σ(i) basis vector. (c) Show that the determinant defines a group homomorphism det : GL n R,whichsendsA det A. (You may use properties of determinants you learned from linear algebra class.) What is the special name we usually give to the kernel of this map? The determinant is a group homomorphism because det(ab) = det(a)det(b) for any two matrices A, B. The identity element of R under multiplication is 1, so the kernel of this map is given by all 6.,

5 Fall 2014 Math 122 Homework Solutions matrices with determinant 1. The special name is the special linear group of rank n, otherwise written as SL n (R). (d) Consider the composite group homomorphism S n GL n R. We call this the sign representation of S n. What is its image? (It is a subgroup of R.) If n 2, the image is the set {±1} = {1, 1} R. By the definition that φ(σ) sendse i to e σ(i), we see that every column of φ(i) has exactly one non-zero entry, given by the number 1. Hence the determinant formula guarantees that det(φ(σ)) is given as some product of many copies of 1 and -1; i.e., the determinant is either 1 or -1. On the other hand, we can see that the permutation (12) S n for any n 2 has determinant -1, for instance by cofactor expansion. Finally, if n = 1, the image is given by the determinant of the identity matrix i.e., the image is the set {1} R. 3. Centers (a) For any group G, thecenter of G is the set of those g such that g commutes with all elements of G. That is, gh = hg for all h. Show that the center of G is a normal subgroup of G. Let Z be the center of G. Note that the identity gh = hg implies that g = hgh 1 for any g Z, h G. (Just multiply by h 1 on the right.) To prove that Z is normal, we must show that for all h G, hzh 1 = Z. Well, for any g Z, we know that hgh 1 = g. Thisshows both Z hzh 1 and hzh 1 Z at the same time; hence Z = hzh 1. (b) What is the center of GL n (R) for n 1? Let A be a matrix in the center. This means that for any invertible matrix B, wehavebab 1 = A. Wenowinterprettheentriesm ij of A as a change-of-basis formula: Let v 1,...,v n be the columns of B. Since B is invertible, these vectors form a basis for R n. Then the entries m ij are numbers such that BAB 1 sends v i to the linear combination n j=1 m ijv j. Importantly, these coefficients m ij are the same for any basis {v i }, because regardless of the basis given by B, wewillhave (m ij )=A. This observation proves that m ij must equal zero for i = j. For if m ij = 0 for some pair i = j, letb be the matrix obtained by scaling the basis vector v j by (for instance) 12, while keeping all other basis vectors the same. This means B defines a new basis v 1 = v 1,..., v j = 12v j,..., v n = v n. 7

6 Fall 2014 Math 122 Homework On the other hand, n BAB 1 (v i )= m ik v j and B AB 1 (v i )= k=1 n n m ik vk = 12m ij v j + m ij v k. Here, we have used the fact that the coefficients m ik are equal in any basis because A is in the center of GL n (R). Moreover, since the matrices BAB 1 and B AB 1 are both equal to A, they define the same linear transformation since v i = vi, and since the v i form a basis, the coefficients above are uniquely determined. Hence we arrive at the statement that 12m ij = m ij. We conclude that m ij =0wheni= j. Now we are home-free. If A must be a diagonal matrix, all of its diagonal entries must be equal, as any B which simply changes the ordering of an ordered basis will swap the eigenvalues of A. QED. Note: There are many ways to prove this face, and I chose what I think is the cleanest. You can also prove that the center of GL n (R) consists of matrices of the form λi by induction on n, or by explicit computations for well-chosen matrices. 4. Using divisibility (a) Let G be a group of order p for some prime p. Let x be a non-identity element of G. Show that x must have order p. The subgroup x generated by x must have order dividing G by Lagrange s theorem. We know x > 1sincex is not the identity. But the only non-negative number that divides p aside from 1 is p because p is prime by assumption; this shows x = p. By definition, the order of x is x, sox has order p. (b) Let G be a group of order p n for some prime p and n 1. Show that G must contain an element of order p. Let x be an element that is not the identity of G. The subgroup g must have some order dividing the order of G by Lagrange s theorem, so we conclude that x pk = 1 for some k between 1 and n. (It is a basic fact that the only numbers dividing p n are numbers of the form p k.) Whatever this k is, consider the element y = x pk 1. By definition, it has order p, sincey p = x p(pk 1) = x pk =1. (c) Let G be a group (possibly infinite). Let H and H be finite subgroups. Show that if gcd( H, H ) = 1, then H H = {1}. If H and H are subgroups of G, soish H. This can be seen as follows: Since H and H are subgroups, 1 G is in both of them. Hence 1 G is in their intersection. Similarly, if both g 1 and g 2 are in both H and 8 k=1 k=i

7 Fall 2014 Math 122 Homework Solutions H, we see that g 1 g 2 is in both H and H by the closure of subgroups. This means g 1 g 2 H H. Finally, if h H and H,thenitsinverseis in both H and H. We conclude that H H is a subgroup of both H and of H. By Lagrange s Theorem, this subgroup must have an order dividing the orders of both H and of H. However, since H and H have greatest common divisor 1, the order of H H must be Using divisibility again (a) Let G be a finite abelian group. Show that the map x x n, for any integer n 0, is a group homomorphism. Let + denote the group operation in G. Thenφ : x x n is in fact the homomorphism x x x, where the addition occurs n times. We then have that φ(x + y) =(x + y)+...+(x + y). One can rearrange the terms on the righthand side because x+y = y+x for an abelian group. We conclude φ(x + y) =(x x)+(y y). This proof did not, in fact, rely on G being finite. (b) Suppose further that gcd( G,n) = 1. Show that the map x x n is a group automorphism of G. Since G is finite, it suffices to show that the map φ : x x n is injective. To show this, it suffices to show that the kernel of φ is trivial. Suppose that x is in the kernel, so that x n = id. Then the group generated by x must have order diving n. 1 At the same time, it is a subgroup of G, so it must have order diving n. Since the only non-negative number dividing both n and G is 1 by hypothesis, x must be a subgroup of G with one element i.e., x must be the identity. This shows that the kernel consists only of the element 1 G. 6. Free groups What does a group with a set of generators, but with no relations look like? If the set of generators is S, this group is called the free group with generating set S. You will prove its existence, and its universal property, in this exercise. 1 To see this, suppose otherwise. Then n = an + r, wheren is the order of x, and where r is between 1 and N 1. We then see that x an+r =(x N ) a x r = x r =1. Hence every element of x is of the form 1,x,...,x r.thisisacontradictionsincer<n and there must be N elements in x by definition. 9

8 Fall 2014 Math 122 Homework Definition 6.1. Let S = {a, b, c,...} be a set. Though I have written a, b, c as though the elements may be enumerable (i.e., countable), S need not be countable. For n 0, a word of length n in S is defined to be a map of sets {1,...,n} S. Theempty word is the map from the empty set to S, and is the unique word of length 0. So a word of length n is simply an ordered string of n elements of S, possibly with repetitions. 2: Example 6.2. If S = {a, b, c}, then here are the words of lengths 0 to (the empty word) a, b, c aa, bb, cc, ab, ba, bc, cb, ca, ac. Here are some examples of words of length 5: aaaba, ababa, acccb. Given a set S, lets be the set given by adjoining a new element for every s S. We will write this new element as s 1 and call it the inverse of s. So for example, if S = {a, b, c,...}, then S = {a, a 1,b,b 1,c,c 1...}. Definition 6.3. A word in S is called reduced if a letter in the word never appears next to its inverse. Remark 6.4. As an example, here are some unreduced words, with unreduced bits underlined. abbaaa 1 bcbcc 1 b, aa 1, ab 1 bb 1 c, ab 1 bb 1 c. Given an unreduced word, we can make it reduced by simply removing two adjacent letters when one is the inverse of the other. For example, here are the reductions of the above words: abbabcbb,, ab 1 c, ab 1 c. Note that to fully reduce a word, one may require a few steps: ab 1 cc 1 ba 1 ab 1 ba 1 aa 1. Regardless, since every word is by definition of finite length, this reduction process terminates. Given any word in S, there is a unique reduction of that word, in which no letter appears next to its inverse. 10

9 Fall 2014 Math 122 Homework Solutions Given two words w 1 and w 2, we may simply concatenate them (i.e., put them side by side) to create a new word. For instance, if then we have w 1 = abc, w 1 w 2 = abcc 1 baa, w 2 = c 1 baa w 2 w 1 = c 1 baaabc. (I ve been told this is a common way to create new words in German.) Note that even if two words are reduced, their concatenation may not be. Also note that w 1 w 2 need not equal w 2 w 1. Definition 6.5. Let S be a set. The free group on S is the set of reduced words of length n 0inS. The group multiplication is given by concatenating two words, then reducing the concatenation. The problem: (a) Show that any word in S admits a unique reduction. (b) Show that the above operation is associative. (c) Show that the free group is in fact a group. (d) Let G be a group, and let j : S G be a map of sets. Show that this extends to a group homomorphism F (S) G. Let s S be an element of the set, and j(s) G its image in G. Let us denote by j : S G the function sending s j(s) and s 1 j(s) 1. We then define a function φ j : Word(S) G by sending any word W = s 1...s l,withs i S, totheelement φ j (s 1 ) φ j (s 2 )... φ j (s l ). We must prove this is well-defined on F (S), and a homomorphism. Well, if w is a reduction of W, it is obtained by canceling pairs of letters appearing next to their inverses. On the other hand, if any letter s appears next to its inverse s 1 inside W, the above string of multiplications in G will also see an appearance of φ j (s) appearing next to φ j (s 1 )=φ j (s) 1. Hence if we cancel two inverse letters in the word W to obtain a new word w, we see that φ j (W )=φ j (w ). More explicitly, given a product of many elements in G, omitting an appearance of φ j (s)φ j (s) 1 (or of φ j (s) 1 φ j (s)) does not change the value of the multiplication: φ j (s 1 )... φ j (s)φ j (s) 1... φ j (s l )) = φ j (s 1 )... 1 G... φ j (s l )) = φ j (s 1 )... φ j (s l )) (and likewise for canceling the appearance of φ j (s) 1 φ j (s)). So if the words wi for i =1,...,I are the words one passes through on reducing 11

10 Fall 2014 Math 122 Homework W to its reduction w, we have a string of equalities: φ j (W )=φ j (w 1)=...= φ j (w I)=φ j (w). This shows that φ j is well-defined on F (S). To show that φ j defines a homomorphism, let W = w w be a concatenation of words, and let w be its reduction. We must prove that φ j (w) =φ j (w ) φ j (w ). By well-definedness, it suffices to show φ j (W )=φ j (w ) φ j (w ). This is obvious, since if then w = s 1...s l, w = s 1...s l φ j (W ) = φ j (s 1)...φ j (s l ) φ j(s 1)... φ j (s l ) = (φ j (s 1)...φ j (s l )) (φ j(s 1)... φ j (s = φ j (w ) φ j (w ). (e) Show there is a bijection of sets {Group homomorphisms F (S) G} = {Set maps S G}. We above define a homomorphism φ j : F (S) G for any function j : S G. This defines a function Φ:{Set maps S G} {Group homomorphisms F (S) G} given by j φ j. We define an inverse map Ψ as follows: If φ is a group homomorphism, it assigns a value to the reduced word s, for any s S. So we define Ψ(φ) :=ψ φ to be the function sending s to φ(s). We must show that Ψ Φ and Φ Ψ are the identities. Well, given a homomorphism φ : F (S) G, letw be the word w = s 1...s l where the s i are whatever letters of S are in w. we know that φ(w) =φ(s 1...s l )=φ(s 1 )... φ(s l ) by the group homomorphism property, and the fact that every word is a product of one-letter words. So the value of φ on one-letters words i.e., its value on S determines its value on all elements of F (S). This shows that Φ Ψ is the identity. On the other hand, we have defined Φ so that Φ(j) =φ j simply sends one-letter words to the value j(s). Hence Ψ Φ is also the identity. This completes the proof. l ) 12

11 Homework 3 1. Cosets of S 3 with respect to S 2 Let S 3 be the symmetric group on 3 elements. Recall that this is the set of all bijections from 3 to itself, where 3 = {1, 2, 3}. Let H S 3 be the set of all bijections τ :3 3 such that τ(3) = 3 i.e., the subset of all bijections that fix 3. (a) Show H is a subgroup of S 3. There are two ways to do this: An explicit and a non-explicit. Explicitly, one sees that the only bijections τ fixing 3 are those of the form 1 G =id 3, and τ = (12). This is because any bijection of 3 fixing 3 must only permute the elements 1, 2 3, and τ is the only permutation of 3 doing this. One sees that τ 2 =id,sotheset{1 G,τ} is a subgroup. Non-explicitly, if τ and τ both fix 3, then τ τ (3) = τ(τ 3) = τ(3) = 3 and τ τ(3) = τ (τ(3) = τ (3) = 3. Hence H is fixed under products. The identity obviously fixed 3, and the inverse of any function with τ(3) = 3 is also a function fixing 3. This shows that H is a subgroup of S 3. (b) So H acts on G = S 3. How many elements are there in the orbit space? That is, how many orbits are there? By the proof of Lagrange s theorem, we know that G is equal to H times the number of orbits of the H action on G. Well, G = 3!, while H has only two elements in it (for instance, by the explicit computation in part (a)). This means that there are exactly 6/2 = 3 orbits. (c) Finally, write out each orbit explicitly. This means you must write out which elements of S 3 are in each orbit. As in a previous homework, we can write the elements of S 3 as id, (12), (23), (13), (123) (132). 13

12 Fall 2014 Math 122 Homework We know that H is one orbit on its own, so we have an orbit given by O 1G = H = {id, (12)}. Arbitrarily choosing an element not in this orbit for instance, (23), we obtain another orbit by looking at (12)(23) and (23): O (23) = {(23), (231)}. Knowing there is only one orbit let, we can put the remaining elements in the third orbit: O 13 = {(13), (321)}. (d) For any n 1, let H S n be the subgroup of all elements that fix n. Exhibit an isomorphism from H to S n 1. Let us define a function φ : S n 1 S n which sends a bijection σ : n 1 n 1 to the bijection φ σ defines as follows: φ σ sends n n, and sends i σ(i) for i = n. This is a group homomorphism because φ σ φ σ (i) =φ σ (σ (i)) = σ(σ (i)) = (σσ )(i) =φ σσ (i) for i = n, and φ σ φ σ (n) =n = φ σσ (n) otherwise. This map is an injection because if φ σ =id n, this means that σ(i) =i for every i. But the only such bijection is the identity bijection, hence the kernel is trivial. It is a surjection because any bijection that fixes n is determined uniquely by a bijection on the remaining elements 1,...,n 1. (e) How many orbits are there of the action of H on S n? As before, the number of orbits is equal to the order of S n divided by the order of H. SinceH is isomorphic to S n 1,thisisn! dividedby (n 1)!. Hence the number of orbits is n. 2. Cyclic groups A group G is called cyclic if there exists g G for which g = G. (a) Show that if two cyclic groups have the same order (finite or otherwise) then they must be isomorphic. Let G and H be two cyclic groups of the same order n. Assume G = g for some g G, and H = h for some h H. Each receives a group homomorphism from Z by sending 1 g, and 1 h. The kernel of both these homomorphisms is nz, while the function is a surjection by the assumption that g and h generate G and H, respectively. Hence by 14

13 Fall 2014 Math 122 Homework Solutions the First Isomorphism Theorem, we have isomorphisms G = Z/nZ = H. By transitivity of group isomorphisms, we are done. (b) Show that S 2 is cyclic. S 2 consists of two elements: the identity, and a non-trivial element τ = (12). τ 2 = 1 so we see that τ generates the whole subgroup. (c) Show that Z is cyclic. Any number n can be written as a finite sum of 1, or of -1. Specifically, we have = n for n positive, where the summation occurs n times. For n negative, we have =n. Hence ZZ = 1. (d) Use Lagrange s theorem to show that any group of prime order must be cyclic. (Hint: Last homework.) Let g = 1, and consider the subgroup g G. This must have order bigger than 1, since it contains at least two distinct elements 1 and g. On the other hand, it must have order dividing G. Since G is prime, the only number bigger than 1 dividing G is G itself. This means g = G, sog = G. (e) Prove that for any integer n 1, there exists a cyclic group of order n. For instance, as a subgroup of S n, or of GL 2 (R), or of C. As a subgroup of S n, one can take the subgroup generated by the cycle σ =(1...n). As a subgroup of GL 2 (R), one can consider the subgroup generated by the matrix of rotating by 2π/n radians. Finally, as a subgroup of C, one can write the subgroup generated by the complex number e 2πi/n. These latter two groups are the easiest to prove as cyclic and of order n clearly, rotation by 2π/n is an operation which becomes the identity after n iterations, so the group generated by this rotation has order at most n. We can enumerate the group elements by how much they rotate the plane: 2π/n, 2(2π/n),...,(n 1)(2π/n). If the order were any smaller than n, we would conclude that a rotation of less than 2π acts as the identity on the plane, which is untrue. 3. Abelian groups A group G is called abelian if for all g 1,g 2 G, wehaveg 1 g 2 = g 2 g 1. (a) Show that S n is not abelian for any n 3. For n = 3, the elements (12) and (123) do not commute,as (12)(123) = (23), (123)(12) = (13). 15

14 Fall 2014 Math 122 Homework These equations hold if we consider the above identities as taking place in S n for any n 3 (i.e., by considering (12), (123), (23), (23) as elements of S n ); hence the two elements do not commute. (b) Show that any cyclic group is abelian. Conclude that S n is not cyclic for any n 3. This is tantamount to showing that the map Z G given by a g a is a group homomorphism. We ll show this fact in (excruciating) detail, only because it is so central to everything. If G = g, any element of G is of the form g a for some a Z. By definition of the notation g a,wehave g a = g... g or g a = g 1... g 1 depending on the sign of a. (In either the case, there are a terms in the product. If a = 0, we take g a = 1.) Then we see case by case that g... g g... g, (a + b terms) = g a g b if a, b 0 g 1... g 1 g 1... g 1,(a + b terms) = g a g b if a, b 0 g... g, (a + b terms) = g... g g 1... g 1, a terms, then b terms = g a g b if a 0,b 0, a + b 0 g a+b g... g, (a + b terms) = = g 1... g 1 g... g, a terms, then b terms = g a g b if a 0,b 0, a + b 0 g 1... g 1,( a + b terms) = g... g g 1... g 1, a terms, then b terms = g a g b if a 0,b 0, a + b 0 g 1... g 1,( a + b terms) = g 1... g 1 g... g, a terms, then b terms = g a g b if a 0,b 0, a + b 0 That is, we have shown that g a+b = g a g b. But a + b = b + a in the integers, so this proves g b g a = g b+a = g a+b = g a g b. Finally, since S n is not abelian for n 3 (see above), it must not be cyclic. (c) Show that the center of an abelian group is the whole group. If a group is abelian, then gh = hg for any pair g, h G. The center of the group is the subgroup Z consisting of those g such that gh = hg for all h G. Hence Z = G if G is abelian. 16

15 Fall 2014 Math 122 Homework Solutions 4. Product groups Let G and H be groups. Define a map m :(G H) (G H) G H, m((g, h), (g,h )) = (gg,hh ). Note that throughout this problem, 1 may refer to either the group unit of G, or the group unit of H. (a) Show that m defines a group structure on G H. The identity is (1 G, 1 H ), as (1 G, 1 H )(g, h) =(1 G g, 1 H h)=(g, h), and (g, h)(1 G, 1 H )=(g1 G,h1 H )=(g, h) for any (g, h) G h. The operation m is associative since G and H have associative operations: (g 1,h 1 )((g 2,h 2 )(g 3,h 3 )) = (g 1,h 1 )(g 2 g 3,h 2 h 3 ) = (g 1 (g 2 g 3 ),h 1 (h 2 h 3 )) = ((g 1 g 2 )g 3, (h 1 h 2 )h 3 ) = (g 1 g 2,h 1 h 2 )(g 3,h 3 ) = ((g 1,h 1 )(g 2,h 2 ))(g 3,h 3 ). Finally, the inverse to (g, h) is given by (g 1,h 1 ), as (g, h)(g 1,h 1 )=(gg 1,hh 1 )=(1 G, 1 H ) and (g 1,h 1 )(g, h) =(g 1 g, h 1 h)=(1 G, 1 H ). (b) Show that (g, 1) (1,h)=(1,h) (g, 1). (g, 1)(1,h)=(g1, 1h) =(g, h) =(1g, h1) = (1,h)(g, 1). (c) Show that if G and H are abelian, then G H is abelian (with the above group structure). (g 1,h 1 )(g 2,h 2 )=(g 1 g 2,h 1 h 2 )=(g 2 g 1,h 2 h 1 )=(g 2,h 2 )(g 1,h 1 ). (d) Show that the maps G G H g (g, 1) 17

16 Fall 2014 Math 122 Homework and G H G, (g, h) g are group homomorphisms. Let f : G G H be the first map. Then f(g 1 g 2 )=(g 1 g 2, 1) = (g 1, 1)(g 2, 1) = f(g 1 )f(g 2 ). Likewise, let p : G H G be the second map. Then p((g 1,h 1 )(g 2,h 2 )) = p((g 1 g 2,h 1 h 2 )) = g 1 g 2 = p((g 1,h 1 )) p((g 2,h 2 )). 18

17 Homework 4 1. Subgroups of Z In this problem, you will show that every subgroup of Z is of the form nz for some n 0. Let H Z be a subgroup which contains some non-zero element. Let n H be the least, positive integer inside H. Show that H = nz. (Hint: Remainders.) Assume there is some k H for which k is not divisible by n. Bythe division algorithm, we know that k = an + r for some integer a and some integer r between 0 and n 1, inclusive. We also conclude r == 0sincek is not divisible by n. On the other hand, if both k and n are in H, then r = k n... n is in H. Sor is some positive integer less than n, but this contradicts the assumption that n is the least positive integer in H. 2. Conjugation actions The conjugation action of a group on itself is by far the most important group action in representation theory. A full understanding of the conjugation action can be illusive, and in many contexts, proves quite essential for research. (a) Fix an element g G. Define a map C g : G G by h ghg 1.Show that C g is a group isomorphism. If h is some element in G,leth = g 1 h g.thenc g (h) =g(g 1 h g)g 1 = 1h 1=h.ThisshowsC g is a surjection. If C g (h) = 1, then ghg 1 = 1, so we conclude h = g 1 g = 1. This means the kernel of C g consists only of 1, meaning C g is injective. We now need only show that C g is a group homomorphism: C g (h 1 h 2 )=g(h 1 h 2 )g 1 = gh 1 g 1 gh 2 g 1 =(gh 1 g 1 )(gh 2 g 1 )=C g (h 1 )C g (h 2 ). Note that I am freely removing and adding parentheses in the products; this is justified by associativity. 19

18 Fall 2014 Math 122 Homework (b) Show that C g C g = C gg. In other words, the assignment g C g defines a group homomorphism G Aut Group (G). So this defines another group action of G on itself. It is quite different from the action we have considered earlier, where all we had was a group homomorphism G Aut Set (G). This new map, G Aut Group (G), is called the conjugation action of G on itself. C g C g (h) =C g (g hg 1 )=g(g hg 1 )g 1 =(gg )h(g 1 g 1 )=(gg )h((gg ) 1 )=C gg (h). Note I m using the property of groups that (ab) 1 = b 1 a 1. (c) If G is abelian, show that C g is trivial for all g G. Since G is abelian, ghg 1 = hgg 1 = h for all g, h. Hence C g (h) = h for all g, h G. Put another way, C g is the identity automorphism of G, for any choice g. 3. Group isomorphisms in general Since C g is a group isomorphism from G to itself, it tells us a lot about the subgroups and elements of G. This is because of some general properties of group isomorphisms, which we now explore. Let φ : G H be a group isomorphism. If K G is a subset, we define φ(k) ={h H such that h = φ(g) for some g K}. (a) Show that isomorphisms preserve orders of elements. That is, show that if g is an element of order n, thenφ(g) is. Note that φ defines a map g to H, simplybysendingg n to φ(g) n. By definition its image is contained in φ(g), and any element inside φ(g) is of the form φ(g) n, so this map is a surjection onto φ(g). Moreover, φ is an isomorphism, so its kernel consists only of 1 G this means the kernel of g H also consists only of 1 G. This proves that g H is an injective group homomorphism with image isomorphic to φ(g). By the first isomorphism theorem, g and φ(g) are isomorphic, hence have equal order. This shows g = φ(g). (b) Show that if K G is a subgroup, it is isomorphic to φ(k). φ defines a homomorphism from K to G, simplybysendingk φ(k). (If you like, this is the composition of the inclusion homomorphism K G with the homomorphism G H.) The kernel of φ consists only of the identity, so the map K H also has trivial kernel. By the first isomorphism theorem, K is isomorphic to the image of K H, which by definition is φ(k). (As an side, note that both (a) and (b) are true simply for injective homomorphisms.) 20

19 Fall 2014 Math 122 Homework Solutions (c) Show that isomorphisms preserve normal subgroups. That is, Show that if K G is a normal subgroup, then φ(k) H is normal. If K G is normal, then for all g G, we know that gkg 1 K. So for any g, wehaveφ(g)φ(k)φ(g 1 )={φ(g)φ(k)φ(g) 1,k K} = {φ(gkg 1 ),k K} = φ(gkg 1 ) φ(k). (This last inclusion follows because if we have subsets A B, theφ(a) φ(b) in general.) On the other hand, since any h H can be written φ(g) for some g G, we see that hφ(k)h 1 = φ(g)φ(k)φ(g) 1 for some g, and hence hφ(k)h 1 φ(k) for all h H. This proves that φ(k) is normal. Note that this only required that φ be a surjective homomorphism. (d) Let K be a normal subgroup G. Show that there is a group isomorphism G/K = H/φ(K). We have a quotient homomorphism H H/φ(K), simply because φ(k) is normal in H. On the other hand, we have the group isomorphism φ : G H. Consider the composition ψ : G H H/φ(K). This is a surjection because both φ and the quotient map are. So by the first isomorphism theorem, G/ ker ψ = H/φ(K). But the kernel of ψ is the set of all g such that φ(g) φ(k); since φ is an injection, this is the set of all g K. That is, the kernel is equal to K. By the first isomorphism theorem, G/K = H/φ(K). Throughout the following exercises, if you have time, think about what the above results imply about elements and subgroups of G that are conjugate. 4. Conjugacy classes of elements (a) Two elements g, g G are called conjugate if there exists some h G such that h 1 gh = g. Show by example that if g and g are conjugate, the choice of h need not be unique. Any element is conjugate to itself; in particular, we could take g = g =1 G ; then any choice of h G satisfies the equation h 1 gh = g. For a less trivial example, consider g = (12),g = (34) inside S 4.Then either of the following choices for h exhibits g as a conjugate of g: For example, h 1 = (1324), (13)(24). (1324)(12)(4231) = (34), (13)(24) ((12)) (13)(24) = (34). 21

20 Fall 2014 Math 122 Homework (b) Show that being conjugate defines an equivalence relation on the set G. That is, show that the relation g g if g is conjugate to g is an equivalence relation. Under this relation, the equivalence class of g is called the conjugacy class of g. g is conjugate to itself by choosing h = 1. If g = h 1 gh, thenlet k = k 1. Then we see g = k 1 g k; hence the relation is symmetric. Finally, if g = x 1 g x,then g = x 1 g x = x 1 h 1 ghx =(hx) 1 ghx. This establishes transitivity. (c) Show that g is the only element in its conjugacy class if and only if g is in the center of G. If g is in the center of G, then for any h G, we have that h 1 gh = h 1 hg = g. Hence the only element conjugate to g is g itself. Conversely, if g is conjugate only to itself, this means that for every h G, wehaveh 1 gh = g. B multiplying both sides of the equation by h on the left, we see that gh = hg. Since this is true for any h G, thisshowsg is in the center of G. 5. Conjugacy classes of subgroups Let H and H be subgroups of G. We say H and H are conjugate if there is some g such that C g (H) =H. That is, if ghg 1 = {ghg 1,h H} = H for some g. (a) Show that being conjugate defines an equivalence relation on the set of all subgroups of G. That is, show that the relation H H if H is conjugate to H is an equivalence relation. The equivalence class of H under this relation is called the conjugacy class of H. Any H is conjugate to itself, since C 1 (H) ={1h1 1 } = {h} = H. This shows reflexivity. If C g (H) =H, then we see that C g 1(H )=H as follows: C g 1(H )={g 1 h g s.t. h H } = {g 1 (ghg 1 )g s.t. h H} = H. The middle equality is using, of course, that any h H is of the form ghg 1 for some h H, and that any element of the form ghg 1 is in H. So we are left to show transitivity: Well, if C g (H) =H and C x (H )=H,thenC x C g (H) =H. (This is just a statement about what functions C do to subsets.) We are finished. (b) Show that H is the only element in its conjugacy class if and only if H is normal. 22

21 Fall 2014 Math 122 Homework Solutions If H is the only element in its conjugacy class, we see that for all g G, C g (H) =ghg 1 = H. This is the definition of being a normal subgroup. Finally, if H is normal, then ghg 1 = H for any g, sothe only subgroup conjugate to H is H itself. 23

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