Review Problems for Final Exam

1. Let Un be the group of multiplicative units mod n (what Goodmanpage 212 calls Φ(n)). The group operation is a ⊗n b = the remainderwhen ab is divided by n. (The equivalence class representatives are theintegers a such that 1 ≤ a ≤ n and g.c.d(a, n) = 1).

(a) Below is a partial multiplication table for U9. Complete the table,then find two different isomorphisms between U9 and (Z6,⊕6),where as usual Z6 = {0, 1, 2, 3, 4, 5}, and ⊕6 is addition mod 6.

Hint: What is the order of 2 in U9?

⊗9 1 2 4 5 7 8

1 1 2 4 5 7 8

2 2 4 8 1 5 7

4 4 8 7 2 1 5

5 5 1 2 7 8 4

7 7 5 1 8 4 2

8 8 7 5 4 2 1

(b) Let Aut(Z9) be set of automorphisms of the group (Z9,⊕9). Inother words, f ∈ Aut(Z9) if and only if f is an isomorphism from(Z9,⊕9) to (Z9,⊕9). With composition of functions as the opera-tion, Aut(Z9) is a cyclic group of order 6. Find an isomorphismbetween U9 and Aut(Z9).

2. Recall that, if the prime factorization of n is n =k∏

k=1

pekk , then Un is

isomorphic to the direct product of the groups Upekk

. On Monday wealso discussed the fact that, if the prime pk is not 2, then Up

ekk

is cyclic,

of order ϕ(pekk ) = pek−1k (pk − 1). (But U2e is not cyclic for e > 2). Youare not responsible for the proofs of these facts (this quarter). However

you should understand and be able to apply these statements. Forexample:

(a) The prime factorization of 191187 is 33 · 73 · 97. Use this fact tofind a direct product of cyclic groups, each of prime power order,such that the direct product is isomorphic to U191187

(b) Find positive integers d1, d2, . . . , dk Such that di is a divisor of di+1

for all i < k. and U191187 is isomorphic to the direct product ofthe groups (Zdi ,⊕di).

(c) Carmichael’s function is defined by λ(n) = the largest of the ordersof the elements in Un. Calculate λ(191187)

3. Recall that, if N is a normal subgroup of a finite group (G, ∗1), then

the quotient group (G/N, ∗2) consists of |G||N | cosets, each of order |N |,and the operation on cosets is defined by

g1N ∗2 g2N = g1∗1g2N

Because N is normal, this operation is well defined (See Theorem 2,page 148 of Pinter). The point of this problem is to understand whatthis means.

(a) If 42016 is divided by 3, what is the remainder?

Hint 4 = 1.

(b) Let G be the symmetric group on [4], i.e. the set of all 4! = 24bijections from [4] to [4], with composition of functions as thegroup operation. Recall that H = {σ ∈ G : σ(4) = 4} is asubgroup of G that is not normal. Let g1 = (1, 2, 3, 4), g∗1 = (1, 4),and g2 = (1, 4). Verify that g1H = g∗1H = {σ : σ(4) = 1}; theyare the same coset (just as 1 = 4 in part a).

(c) Since g1H and g∗1H are just two different ways to write the samecoset, they should give the same answer when multiplied by thecoset g2H. Verify that, in fact, g1g2H 6= g∗1g2H, so multiplicationof cosets is not well-defined.

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4. Let G be the automorphism group of the above graph above, i.e. thesubgroup of S7 that preserves adjacency. Let H be the subgroup ofG consisting of permutations that fix at least 5, 6 and 7. Thus H ={ε, (1, 2)}.

(a) Write down all six left cosets of H in the same format, as sets oftwo permutations:

• ε = εH = H = {ε, (1, 2)}.• (5, 6) = (5, 6)H = {(5, 6), (5, 6)(1, 2)}• (5, 7) = (5, 7)H =

• (6, 7) = (6, 7)H =

• (5, 6, 7) = (5, 6, 7)H =

• (5, 7, 6) = (5, 7, 6)H =

(b) Complete the multiplication table for the quotient space G/H:

◦ ε (5, 6) (5, 7) (6, 7) (5, 6, 7) (5, 7, 6)

ε ε (5, 6) (5, 7) (6, 7) (5, 6, 7) (5, 7, 6)

(5, 6) (5, 6)

(5, 7) (5, 6)

(6, 7) (5, 6)

(5, 6, 7) (5, 7, 6) ε

(5, 7, 6)

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(c) Find an isomorphism between G/H and S3

5. If H and K are subgroups of a group G, then |HK| = |H||K||H∩K| . If K is

normal, then HK is a subgroup.

6. Recall that the order of an element is the order of the cyclic subgroupthat it generates: order(a) = min{k ∈ Z+ : ak = ε} For abelian groupsit is conventional to use additive notation:

order(a) = min{k ∈ Z+ : a+ a+ . . . a︸ ︷︷ ︸ktimes

= 0},

where 0 is the additive identity element of the group (not necessarilyan integer).

(a) Show that, for 1 ≤ a ≤ n, the order of a in (Zn,⊕n) is ng.c.d.(n,a)

(b) For any choice of m and n, there is a “trivial”homomorphismfrom (Zm,⊕m) to (Zn,⊕n) defined by f(x) = 0 for all x. Showthat there is no non-trivial homomorphism from (Z331,⊕m) to(Z2016,⊕n).

Hint: 331 and 2016 are relatively prime.

7. For each of the following pairs of groups, determine whether G1 andG2 are isomorphic.

(a) G1 = the group of upper triangular 2× 2 matrices having 1’s thediagonal, and arbitrary elements of the ring Z5 above the diagonal,G2 = Z5.

(b) G1 = Z3 × Z5, G2 = Z15,

(c) G1 = Z3 × Z3, G2 = Z6

(d) G1 = Z2 × Z5, G2 = the dihedral group of order 10 consisting ofthe symmetries of a regular pentagon (generated by a reflectionand a rotation of 360

5degrees.)

8. If S ⊆ G let 〈S〉 denote the subgroup generated by S. One way todescribe 〈S〉 is that it is intersection of all the subgroups of G thatcontain S. A more constructive way, is to say 〈S〉 is the subgroupconsisting of all elements of G that can be written as a product ofelements of S. For each of the following choices of G and S, find thesubrgroup 〈S〉.

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(a) G = Z30, S = {6, 10}(b) G = Z4 × Z4 , S = {(1, 1)}(c) G = Z4 × Z4 , S = {(1, 1), (0, 1)}(d) G= the group of 24 symmetries of the cube that we have consid-

ered previously, S = {Rz} where Rz = (1, 2, 3, 4)(5, 6, 7, 8), a 90degree rotation about the z-axis.

(e) G= the group of 24 symmetries of the cube that we have consid-ered previously, S = {Rx, Rz} where Rx = (5, 1, 2, 6)(8, 4, 3, 7) isa 90 degree rotation about the x-axis,

9. Suppose G1, G2 are groups with operations ∗i and identity elementsεi, i = 1, 2. Which of the following must be true if ϕ : G1 :→ G2 is ahomomorphism.

(a) |G1| ≥ |ϕ(G1)|(b) |G1| is divisible by |ϕ(G1)|.(c) For every a ∈ G1, the order of a is divisible by the order of ϕ(a)

in G2.

(d) ϕ(G1) is a subgroup of G2.

(e) G1/Kerφ is isomorphic to a subgroup of G2.

(f) |G1| is divisible by |ϕ(G1)|(g) ϕ(ε1) = ϕ(ε2)

(h) If G1 is abelian, then G2 is abelian.

(i) If G2 is abelian, then G1 is abelian.

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