Please show all work! Answers without supporting work will not be given credit.

Name: ! ! ! ! ! ! ! ! ! Number:

1. (18 points)

a) Complete the following definitions.

! ai) Let G and G be groups with respect to binary operations ∗ and ∗ , respectively. A mapping φ :G→G is a homomorphism if φ x∗ y( ) = φ x( )∗φ y( ) for all x, y∈G .

! aii) If φ :G→G is a group homomorphism then the kernel of φ is

kerφ = x ∈G :φ x( ) = e{ } where e is the identity element of G .

b) Label each of the following statements as either true or false. Justify your answers.

! bi) The empty set ∅ is a subgroup of any group G . True / False. Reason: Any subgroup contains an identity element but the empty set has no element.

! bii) If a subgroup H of a group G is abelian, then G must be abelian.! True / False. Reason: H = 1( ), 12( ){ } is an abelian subgroup of the group S3 but S3

is not abelian for all x, y∈S3 because 123( ) 12( ) = 13( ) ≠ 23( ) = 12( ) 123( ) .

c) Give an example.

! ci) Find two groups G and G such that G is a homomorphic image of G but G is not a homomorphic image of G . Example: Let G = and G

= n . Define φ :→ n by φ a( ) = a[ ] . Then φ is an epimorphism so n is a homomorphic image of . But is not a homomorphic image of

n since there does exist no surjective function from n to .

! cii) Find a subset of that is closed under addition but is not a subgroup of the additive group . ! Example: + is a subset of that is closed under addition but + is not a subgroup of .

MTM 0522062 MIDTERM EXAM I! APRIL 16, 2013

NAME: NUMBER: ! 19:00 - 20:30

! ALGEBRA! 1

2. (10 points)

For a fixed element a of a multiplicative group G , the set Ca = x ∈G :ax = xa{ } is called the centralizer of a in G . Prove that for any a∈G , Ca is a subgroup of G.

Exam II MAT 416 Spring 2013

Name:Directions: Please show all work and be as neat as possible.

1.18 points Complete the following definitions.a) An element a of a group G is a generator for G if

G = hai.

b) Let G and G0 be groups with respect to binary operations ⇤ and ⇤0 respectively. A mapping f : G ! G0 is ahomomorphism iff(x⇤ y) = f(x)⇤0 f(y) for all x,y 2 G.

c) A mapping f : G ! G0 is an isomorphism iff is a homomorphism that is one-to-one and onto.

d) If f : G ! G0 is a group homomorphism then the kernel of f iskerf = {x 2 G | f(x) = e0} where e0 is the identity element of G0.

e) If f : G ! G0 is a group homomorphism then the image of f isimf = {y 2 G0 | there exists x 2 G with f(x) = y}.

f) If G is a group and x 2 G then the order of x isthe number of elements in the cyclic subgroup generated by x.

2.10 points For a fixed element a of a multiplicative group G, the set Ca = {x 2 G | ax = xa} is called the centralizer of ain G. Prove that for any a 2 G, Ca is a subgroup of G.

Proof. First let e be the identity element of G, and observe that ae = ea. Thus e 2Ca. Hence Ca 6= /0.Next let x,y 2Ca. Then

a(xy) = (ax)y by associativity= (xa)y as x 2Ca

= x(ay) by associativity= x(ya) as y 2Ca

= (xy)a by associativity

Thus xy 2Ca.Finally, let x 2Ca. Then

ax = xa=) x�1ax = a multiplying by x�1 on the left=) x�1a = ax�1 multiplying by x�1 on the right

Thus x�1 2Ca.Therefore Ca is a subgroup of G.

3.10 points Suppose G = hai is a cyclic group of order 14.a) List all generators of G.

a,a3,a5,a9,a11,a13

b) List all distinct subgroups of G and their elements.hai= {a0,a,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13}= G⌦a2↵= {a0,a2,a4,a6,a8,a10,a12}⌦

a7↵= {a0,a7}⌦

a14↵= {a0}

MTM 0522062 MIDTERM EXAM I! APRIL 16, 2013

NAME: NUMBER: ! 19:00 - 20:30

! ALGEBRA! 2

3. (10 points)

Suppose G = a is a cyclic group of order 14.

a) List all generators of G.

Exam II MAT 416 Spring 2013

Name:Directions: Please show all work and be as neat as possible.

1.18 points Complete the following definitions.a) An element a of a group G is a generator for G if

G = hai.

b) Let G and G0 be groups with respect to binary operations ⇤ and ⇤0 respectively. A mapping f : G ! G0 is ahomomorphism iff(x⇤ y) = f(x)⇤0 f(y) for all x,y 2 G.

c) A mapping f : G ! G0 is an isomorphism iff is a homomorphism that is one-to-one and onto.

d) If f : G ! G0 is a group homomorphism then the kernel of f iskerf = {x 2 G | f(x) = e0} where e0 is the identity element of G0.

e) If f : G ! G0 is a group homomorphism then the image of f isimf = {y 2 G0 | there exists x 2 G with f(x) = y}.

f) If G is a group and x 2 G then the order of x isthe number of elements in the cyclic subgroup generated by x.

2.10 points For a fixed element a of a multiplicative group G, the set Ca = {x 2 G | ax = xa} is called the centralizer of ain G. Prove that for any a 2 G, Ca is a subgroup of G.

Proof. First let e be the identity element of G, and observe that ae = ea. Thus e 2Ca. Hence Ca 6= /0.Next let x,y 2Ca. Then

a(xy) = (ax)y by associativity= (xa)y as x 2Ca

= x(ay) by associativity= x(ya) as y 2Ca

= (xy)a by associativity

Thus xy 2Ca.Finally, let x 2Ca. Then

ax = xa=) x�1ax = a multiplying by x�1 on the left=) x�1a = ax�1 multiplying by x�1 on the right

Thus x�1 2Ca.Therefore Ca is a subgroup of G.

3.10 points Suppose G = hai is a cyclic group of order 14.a) List all generators of G.

a,a3,a5,a9,a11,a13

b) List all distinct subgroups of G and their elements.hai= {a0,a,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13}= G⌦a2↵= {a0,a2,a4,a6,a8,a10,a12}⌦

a7↵= {a0,a7}⌦

a14↵= {a0}

b) List all distinct subgroups of G and their elements.

Exam II MAT 416 Spring 2013

Name:Directions: Please show all work and be as neat as possible.

1.18 points Complete the following definitions.a) An element a of a group G is a generator for G if

G = hai.

b) Let G and G0 be groups with respect to binary operations ⇤ and ⇤0 respectively. A mapping f : G ! G0 is ahomomorphism iff(x⇤ y) = f(x)⇤0 f(y) for all x,y 2 G.

c) A mapping f : G ! G0 is an isomorphism iff is a homomorphism that is one-to-one and onto.

d) If f : G ! G0 is a group homomorphism then the kernel of f iskerf = {x 2 G | f(x) = e0} where e0 is the identity element of G0.

e) If f : G ! G0 is a group homomorphism then the image of f isimf = {y 2 G0 | there exists x 2 G with f(x) = y}.

f) If G is a group and x 2 G then the order of x isthe number of elements in the cyclic subgroup generated by x.

2.10 points For a fixed element a of a multiplicative group G, the set Ca = {x 2 G | ax = xa} is called the centralizer of ain G. Prove that for any a 2 G, Ca is a subgroup of G.

Proof. First let e be the identity element of G, and observe that ae = ea. Thus e 2Ca. Hence Ca 6= /0.Next let x,y 2Ca. Then

a(xy) = (ax)y by associativity= (xa)y as x 2Ca

= x(ay) by associativity= x(ya) as y 2Ca

= (xy)a by associativity

Thus xy 2Ca.Finally, let x 2Ca. Then

ax = xa=) x�1ax = a multiplying by x�1 on the left=) x�1a = ax�1 multiplying by x�1 on the right

Thus x�1 2Ca.Therefore Ca is a subgroup of G.

3.10 points Suppose G = hai is a cyclic group of order 14.a) List all generators of G.

a,a3,a5,a9,a11,a13

b) List all distinct subgroups of G and their elements.hai= {a0,a,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13}= G⌦a2↵= {a0,a2,a4,a6,a8,a10,a12}⌦

a7↵= {a0,a7}⌦

a14↵= {a0}

MTM 0522062 MIDTERM EXAM I! APRIL 16, 2013

NAME: NUMBER: ! 19:00 - 20:30

! ALGEBRA! 3

4. (10 points)

Let G = 11 − 0[ ]{ } under multiplication, which is a cyclic group.a) List all generators of G.

Exam II MAT 416 Spring 20134.10 points Let G = Z11 �{[0]} under multiplication, which is a cyclic group.

a) List all generators of G.[2], [8], [7], [6]

b) List all distinct subgroups of G and their elements.h[2]i= Gh[4]i= {[1], [4], [5], [9], [3]}h[10]i= {[1], [10]}h[1]i= {[1]}

5.10 points a) Let G be a multiplicative group and a 2 G. If o(a) = mn for some m,n 2 Z+ prove that o(am) = n.

Proof. First observe that (am)n = amn = e as o(a) = mn. Thus o(am) n.Now assume (am)k = e for some k 2 Z with 0 < k < n. Then amk = (am)k = e, and 0 < nk < mn.However, this contradicts that o(a) = mn, which means that al 6= e for any l 2 Z with 0 < l < mn. Thuso(am) = n.

b) Let G = Z23 �{[0]} under multiplication. Given that [5] is a generator for G, what is o([2])?Note that o([5]) = |h[5]i |= |G|= 22, with the middle equality following from the fact that [5] is a generatorfor G. Also note that [5]2 = [2]. Thus o([2]) = o([5]2) = 11 by part a.

6.10 points Consider the additive groups Z and Z20. Define f : Z! Z20 by f(x) = [5x]. Prove that f is a homomorphismand find kerf.

Proof. Let x,y 2 Z. Then f(x+ y) = [5(x+ y)] = [5x+ 5y] = [5x] + [5y] = f(x) + f(y). Therefore f is ahomomorphism.Now x 2 kerf () [5x] = [0] () 5x ⌘ 0 (mod 20) () 5x = 20n for some n 2 Z () x = 4n for somen 2 Z. Thus kerf = {4n | n 2 Z}.

Page 2

b) List all distinct subgroups of G and their elements.

Exam II MAT 416 Spring 20134.10 points Let G = Z11 �{[0]} under multiplication, which is a cyclic group.

a) List all generators of G.[2], [8], [7], [6]

b) List all distinct subgroups of G and their elements.h[2]i= Gh[4]i= {[1], [4], [5], [9], [3]}h[10]i= {[1], [10]}h[1]i= {[1]}

5.10 points a) Let G be a multiplicative group and a 2 G. If o(a) = mn for some m,n 2 Z+ prove that o(am) = n.

Proof. First observe that (am)n = amn = e as o(a) = mn. Thus o(am) n.Now assume (am)k = e for some k 2 Z with 0 < k < n. Then amk = (am)k = e, and 0 < nk < mn.However, this contradicts that o(a) = mn, which means that al 6= e for any l 2 Z with 0 < l < mn. Thuso(am) = n.

b) Let G = Z23 �{[0]} under multiplication. Given that [5] is a generator for G, what is o([2])?Note that o([5]) = |h[5]i |= |G|= 22, with the middle equality following from the fact that [5] is a generatorfor G. Also note that [5]2 = [2]. Thus o([2]) = o([5]2) = 11 by part a.

6.10 points Consider the additive groups Z and Z20. Define f : Z! Z20 by f(x) = [5x]. Prove that f is a homomorphismand find kerf.

Proof. Let x,y 2 Z. Then f(x+ y) = [5(x+ y)] = [5x+ 5y] = [5x] + [5y] = f(x) + f(y). Therefore f is ahomomorphism.Now x 2 kerf () [5x] = [0] () 5x ⌘ 0 (mod 20) () 5x = 20n for some n 2 Z () x = 4n for somen 2 Z. Thus kerf = {4n | n 2 Z}.

Page 2

MTM 0522062 MIDTERM EXAM I! APRIL 16, 2013

NAME: NUMBER: ! 19:00 - 20:30

! ALGEBRA! 4

5. (10 points)

a) Let a be element of a finite group G . Prove that a and a−1 (the inverse of a ) have the same order.

f) Let G = a be a cyclic group of order 13. The divisors of 13 are 1 and 13 so the distinct subgroups of G are

a = G,

a13 = a13 = e{ }.

23. Since o a( ) = 24 , a24 = e .

a) 2 is the least positive integer such that a12( )2 = a24 = e , so o a12( ) = 2.b) a8,a16

c) a6,a18

d) none

24. Since o a( ) = 35 , a35 = e . a) noneb) a7,a14 ,a21,a28

c) a5,a10,a15,a20,a25,a30

d) none

25. If H ⊂ is a subgroup, then either H = 0{ } is trivial or there exists a unique integer n ≥1 such that H = n = nk : k ∈{ } .

26. All generators of an infinite cyclic group are a and a−1 , so G = a = a−1 . For example,

the additive group is cyclic and = 1 = −1 .

27. First note that a = bn for some n∈ since a∈ b . Now let x ∈ a . This means that

x = am for some m∈ . Thus x = am = bn( )m = bnm ∈ b . Since this is true for all x ∈ a we

have a ⊆ b .

28. a)

ak( )−1 = a−k = a−1( )kak = e

⇒ a−1( )k = e−1 = e⇒ o a−1( ) ≤ o a( ) (1)

Similarly,

Exercises 3.4" Cyclic Groups

19

a−1( )k = e⇒ a−1( )−1( )k = e−1⇒ ak = e−1 = e

⇒ o a( ) ≤ o a−1( ) (2)

By (1) and (2), o a( ) = o a−1( ).

b)

Section 3.4 Solutions 2

3.4.28 Let a and b be elements of a finite group G.b. Prove that a and bab

�1 have the same order.

c. Prove that ab and ba have the same order.

Proof of b. Let e be the identity element of G. First observe that for any x,y 2 G and for any integer m with m �1, (xyx

�1)m = xy

m

x

�1. To prove this proceed by induction on n. For n = 1 we clearly have (xyx

�1)1 = xy

1x

�1.So assume that (xyx

�1)k = xy

k

x

�1 for some integer k � 1. Then(xyx

�1)k+1 = (xyx

�1)k(xyx

�1)= (xy

k

x

�1)(xyx

�1) by the inductive hypothesis= xy

k

eyx

�1

= xy

k

yx

�1

= xy

k+1x

�1. This completes the induction.

Now let n = o(a). This means that a

n = e and n is the least such positive integer. Then we have (bab

�1)n =ba

n

b

�1 = beb

�1 = bb

�1 = e. Thus o(bab

�1) n.Next assume that (bab

�1)m = e for some integer m with 0 < m < n. Then e = (bab

�1)m = ba

m

b

�1. Nowba

m

b

�1 = e

=) a

m

b

�1 = b

�1 multiplying by b

�1 on the left=) a

m = e multiplying by b on the right

However, this contradicts that n is the least integer such that a

n = e. Therefore o(bab

�1) = n = o(a).

Proof of c. First observe that for any x,y 2 G, (xy)m = x(yx)m�1y for all integers m � 1. To prove this proceed

by induction on n. For m = 1 we have (xy)1 = xy = xey = x(yx)0y. So assume that (xy)k = x(yx)k�1

y for someinteger k � 1. Then

(xy)k+1 = (xy)k(xy)= x(yx)k�1

y(xy) by the inductive hypothesis= x(yx)k�1(yx)y by associativity= x(yx)k

y This completes the induction.

Now let n = o(ab). This means that (ab)n = e and n is the least such positive integer. So(ab)n = e

=) a(ba)n�1b = e by the above observation

=) (ba)n�1b = a

�1 multiplying by a

�1 on the left=) (ba)n�1

ba = e muliplying by a on the right=) (ba)n = e.

Thus o(ba) n.Next assume that (ba)m = e for some integer m with 0 < m < n. Then

(ba)m = e

=) b(ab)m�1a = e by the above observation

=) (ab)m�1a = b

�1 multiplying by b

�1 on the left=) (ab)m�1

ab = e multiplying by b on the right=) (ab)m = e.

However, this contradicts that n is the least integer such that (ab)n = e. Therefore o(ba) = n = o(ab).

c)

Section 3.4 Solutions 2

3.4.28 Let a and b be elements of a finite group G.b. Prove that a and bab

�1 have the same order.

c. Prove that ab and ba have the same order.

Proof of b. Let e be the identity element of G. First observe that for any x,y 2 G and for any integer m with m �1, (xyx

�1)m = xy

m

x

�1. To prove this proceed by induction on n. For n = 1 we clearly have (xyx

�1)1 = xy

1x

�1.So assume that (xyx

�1)k = xy

k

x

�1 for some integer k � 1. Then(xyx

�1)k+1 = (xyx

�1)k(xyx

�1)= (xy

k

x

�1)(xyx

�1) by the inductive hypothesis= xy

k

eyx

�1

= xy

k

yx

�1

= xy

k+1x

�1. This completes the induction.

Now let n = o(a). This means that a

n = e and n is the least such positive integer. Then we have (bab

�1)n =ba

n

b

�1 = beb

�1 = bb

�1 = e. Thus o(bab

�1) n.Next assume that (bab

�1)m = e for some integer m with 0 < m < n. Then e = (bab

�1)m = ba

m

b

�1. Nowba

m

b

�1 = e

=) a

m

b

�1 = b

�1 multiplying by b

�1 on the left=) a

m = e multiplying by b on the right

However, this contradicts that n is the least integer such that a

n = e. Therefore o(bab

�1) = n = o(a).

Proof of c. First observe that for any x,y 2 G, (xy)m = x(yx)m�1y for all integers m � 1. To prove this proceed

by induction on n. For m = 1 we have (xy)1 = xy = xey = x(yx)0y. So assume that (xy)k = x(yx)k�1

y for someinteger k � 1. Then

(xy)k+1 = (xy)k(xy)= x(yx)k�1

y(xy) by the inductive hypothesis= x(yx)k�1(yx)y by associativity= x(yx)k

y This completes the induction.

Now let n = o(ab). This means that (ab)n = e and n is the least such positive integer. So(ab)n = e

=) a(ba)n�1b = e by the above observation

=) (ba)n�1b = a

�1 multiplying by a

�1 on the left=) (ba)n�1

ba = e muliplying by a on the right=) (ba)n = e.

Thus o(ba) n.Next assume that (ba)m = e for some integer m with 0 < m < n. Then

(ba)m = e

=) b(ab)m�1a = e by the above observation

=) (ab)m�1a = b

�1 multiplying by b

�1 on the left=) (ab)m�1

ab = e multiplying by b on the right=) (ab)m = e.

However, this contradicts that n is the least integer such that (ab)n = e. Therefore o(ba) = n = o(ab).

Exercises 3.4" Cyclic Groups

20

b) Let G = 23 − 0[ ]{ } under multiplication. Given that 5[ ] is a generator for G , what is

o 14[ ]( ) = ?Note that o 5[ ]( ) = o G( ) = 22 with the middle equality following from the fact 5[ ] is a

generator for G . 5[ ]−1 is also a generator for G . Since 5[ ]−1 = 14[ ] , by part a,

o 5[ ]( ) = o 14[ ]( ) = 22.

MTM 0522062 MIDTERM EXAM I! APRIL 16, 2013

NAME: NUMBER: ! 19:00 - 20:30

! ALGEBRA! 5

6. (10 points)

Consider the additive groups and 20 . Define φ :→ 20 by φ x( ) = 5x[ ] . Prove that φ is a homomorphism and find kerφ .

Exam II MAT 416 Spring 20134.10 points Let G = Z11 �{[0]} under multiplication, which is a cyclic group.

a) List all generators of G.[2], [8], [7], [6]

b) List all distinct subgroups of G and their elements.h[2]i= Gh[4]i= {[1], [4], [5], [9], [3]}h[10]i= {[1], [10]}h[1]i= {[1]}

5.10 points a) Let G be a multiplicative group and a 2 G. If o(a) = mn for some m,n 2 Z+ prove that o(am) = n.

Proof. First observe that (am)n = amn = e as o(a) = mn. Thus o(am) n.Now assume (am)k = e for some k 2 Z with 0 < k < n. Then amk = (am)k = e, and 0 < nk < mn.However, this contradicts that o(a) = mn, which means that al 6= e for any l 2 Z with 0 < l < mn. Thuso(am) = n.

b) Let G = Z23 �{[0]} under multiplication. Given that [5] is a generator for G, what is o([2])?Note that o([5]) = |h[5]i |= |G|= 22, with the middle equality following from the fact that [5] is a generatorfor G. Also note that [5]2 = [2]. Thus o([2]) = o([5]2) = 11 by part a.

6.10 points Consider the additive groups Z and Z20. Define f : Z! Z20 by f(x) = [5x]. Prove that f is a homomorphismand find kerf.

Proof. Let x,y 2 Z. Then f(x+ y) = [5(x+ y)] = [5x+ 5y] = [5x] + [5y] = f(x) + f(y). Therefore f is ahomomorphism.Now x 2 kerf () [5x] = [0] () 5x ⌘ 0 (mod 20) () 5x = 20n for some n 2 Z () x = 4n for somen 2 Z. Thus kerf = {4n | n 2 Z}.

Page 2

MTM 0522062 MIDTERM EXAM I! APRIL 16, 2013

NAME: NUMBER: ! 19:00 - 20:30

! ALGEBRA! 6

7. (10 points)

Consider the subgroup G = R0,R90,R180,R270{ } of the group of rigid motions of the square,

and the group G = 5 − 0[ ]{ } under multiplication. Define an isomorphism from G to G .

Define φ :G→G by φ R0( ) = 1[ ],φ R90( ) = 2[ ],φ R180( ) = 4[ ], and φ R270( ) = 3[ ]. Clearly φ is one-to-one and onto. The following tables demonstrate that φ is a homomorphism.

R0 R90 R180 R270

R0

R90

R180

R270

R0 R90 R180 R270

R90 R180 R270 R0

R180 R270 R0 R90

R270 R0 R90 R180

↓φ

× 1[ ] 2[ ] 4[ ] 3[ ]

1[ ]

2[ ]

4[ ]

3[ ]

1[ ] 2[ ] 4[ ] 3[ ]

2[ ] 4[ ] 3[ ] 1[ ]

4[ ] 3[ ] 1[ ] 2[ ]

3[ ] 1[ ] 2[ ] 4[ ]

MTM 0522062 MIDTERM EXAM I! APRIL 16, 2013

NAME: NUMBER: ! 19:00 - 20:30

! ALGEBRA! 7

8. (10 points)

For a fixed element a in a group G , define a mapping ta :G→G by ta x( ) = axa−1 . Prove that ta is an automorphism of G .

Exam II MAT 416 Spring 20137.10 points Consider the subgroup G = {R0,R90,R180,R270} of the group of symmetries of the square, and the group G0 =

Z5 �{[0]} under multiplication. Define an isomorphism from G to G0. Give the matching multiplication tablesthat demonstrate that this is an isomorphism.

Define f : G ! G0 by f(R0) = [1], f(R90) = [2], f(R180) = [4], and f(R270) = [3]. Clearly f is one-to-one andonto. The following tables demonstrate that f is a homomorphism.

· R0 R90 R180 R270

R0 R0 R90 R180 R270R90 R90 R180 R270 R0R180 R180 R270 R0 R90R270 R270 R0 R90 R180

f//

+ [1] [2] [4] [3][0] [1] [2] [4] [3][2] [2] [4] [3] [1][4] [4] [3] [2] [1][3] [3] [2] [1] [4]

Therefore f is an isomorphism.

8.10 points For a fixed element a in a group G, define a mapping ta : G ! G by ta(x) = axa�1. Prove that ta is anautomorphism of G.

Proof. Let x,y 2 G. Then ta(xy) = axya�1 = axeya�1 = (axa�1)(aya�1) = ta(x)ta(y). Thus ta is ahomomorphism.Now assume that ta(x) = e. Then e = ta(x) = axa�1. Thus

e = axa�1

=) a = ax multiplying by a on the right=) e = x multiplying by a�1 on the left

Thus ker(ta) = {e}. Hence ta is one-to-one.Finally, let x 2 G. Letting y = a�1xa we see ta(y) = ta(a�1xa) = a(a�1xa)a�1 = x. Thus ta is onto.Therefore ta is an isomorphism.

9.12 points Let f = (1,3,5,4,2)(1,4,3,5) and g = (1,3,6)(2,4,5).

a) Express f as a product of disjoint cycles. (1,2)(3,4,5)

b) f �g = (1,4,3,6,2,5)

c) f 3 = (1,2)

d) f�1 = (1,2)(3,5,4)

Page 3

MTM 0522062 MIDTERM EXAM I! APRIL 16, 2013

NAME: NUMBER: ! 19:00 - 20:30

! ALGEBRA! 8

9. (12 points)

Let f = 1,3,5,4,2( ) 1,4,3,5( ) and g = 1,3,6( ) 2,4,5( ) . a) Express f as a product of disjoint cycles:b) f g = 1,4,3,6,2,5( )c) f 3 = 1,2( )d) f −1 = 1,2( ) 3,5,4( )

MTM 0522062 MIDTERM EXAM I! APRIL 16, 2013

NAME: NUMBER: ! 19:00 - 20:30

! ALGEBRA! 9