Math 430 – Problem Set 2 Solutions

Due February 5, 2016

3.1(e). Find all x ∈ Z satisfying 5x ≡ 1 (mod 6)

Solution. In Z6 , 5 + 6Z is its own inverse. Multiplying both sides by 5 yields x ≡ 5 (mod 6).

3.1(f). Find all x ∈ Z satisfying 3x ≡ 1 (mod 6)

Solution. The multiples of 3 modulo 6 are 0 and 3, so there are no solutions to this equation.

3.7. Let S = R\{−1} and define a binary operation on S by a ∗ b = a + b + ab. Prove that (S, ∗) is an
abelian group.
Solution.

• We first show that the operation gives a function S × S → S. Certainly a ∗ b ∈ R, so we just need
to show that if a, b ∈ S then a ∗ b 6= −1. If a ∗ b = −1 then 1 + a + b + ab = 0, or (1 + a)(1 + b) = 0.
This is impossible since a 6= −1 and b 6= −1.
• We show that 0 is the identity for S: for any a ∈ S, we have 0 ∗ a = 0 + a + 0 · a = a = a + 0 + a · 0 =
a ∗ 0.
• We show that the operation is associative:

a ∗ (b ∗ c) = a ∗ (b + c + bc)
= a + b + c + bc + a(b + c + bc)
= a + b + c + bc + ab + ac + abc
= a + b + ab + c + (a + b + ab)c
= (a + b + ab) ∗ c
= (a ∗ b) ∗ c.

−a −a
• We show that if a ∈ S then 1+a ∈ S is its inverse. Note that 1+a ∈ R since a 6= −1. Moreover, if
−a −a
1+a = −1 then −a = −1 − a, which is impossible. Thus 1+a ∈ S. We then compute

−a −a −a2
a∗ =a+ + =0
1+a 1+a 1+a
−a −a −a2
∗a= +a+ =0
1+a 1+a 1+a

• Finally, note that a ∗ b = a + b + ab = b ∗ a since addition and multiplication are commutative in

R.
Thus (S, ∗) is an abelian group.

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3.16. Give a specific example of some group G and elements g, h ∈ G where (gh)n 6= g n hn .
Solution. For n = 2, any g, h with gh 6= hg will work. For example, in S3 we have

[(12)(13)]2 = (132)2
= (123)
(12)2 (13)2 = () · ()
= ().

3.17. Give an example of three different groups with eight elements. Why are the groups different?
Solution. There are five groups of order eight, up to isomorphism: you can select any three. They
are
• Z8 ,
• Z4 × Z2 ,
• Z2 × Z2 × Z2 ,
• D4 ,
• Q8 .
The first three are abelian, and thus different from the last two. The first three are distinguished from
each other by the largest order of an element (8 vs 4 vs 2). To see that D4 and Q8 are not isomorphic,
note that D4 has four elements of order 2 (the four reflections) while Q8 only has one (−1).

3.22. Show that addition and multiplication mod n are well defined operations. That is, show that the
operations do not depend on the choice of the representative from the equivalence classes mod n.
Solution. Suppose that a ≡ b (mod n) and c ≡ d (mod n). Then there are integers r, s with a = b+rn
and c = d + sn. We find that

a + c = b + rn + d + sn
= b + d + (r + s)n,

so a + c ≡ b + d (mod n) and thus addition is well defined. Similarly,

ac = (b + rn)(c + sn)
= bc + bsn + crn + rsn2
= bc + (bs + cr + rsn)n,

so ac ≡ bd (mod n) and thus multiplication is well defined.

3.25. Let a and b be elements in a group G. Prove that abn a−1 = (aba−1 )n for n ∈ Z.
Solution.
• For n = 0, this is the statement that a · 1 · a−1 = (aba−1 )0 , which is true since both sides are the
identity.
• For n > 0 we prove the statement by induction. Suppose that abn−1 a−1 = (aba−1 )n−1 . Then

(aba−1 )n = (aba−1 )n−1 (aba−1 )

= abn−1 a−1 aba−1
= abn a−1 .

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 • Finally, for n < 0, let m = −n. Using the statement for m > 0, we have

(aba−1 )n = ((aba−1 )−1 )m

= (ab−1 a−1 )m
= a(b−1 )m a−1
= abn a−1

3.31. Show that if a2 = e for all elements a in a group G then G must be abelian.
Solution. Suppose a, b ∈ G. Then e = (ab)(ab) and e = (ab)(ba) since b2 = e and a2 = e. Since
inverses are unique, ab = ba. Thus G is abelian.

3.33. Let G be a group and suppose that (ab)2 = a2 b2 for all a and b in G. Prove that G is an abelian group.
Solution. For all a, b ∈ G we have
abab = aabb.
Multiplying on the left by a−1 and on the right by b−1 yields ba = ab, so G is abelian.

3.40. Let n o
cos(θ) − sin(θ)
G= sin(θ) cos(θ)
,

where θ ∈ R. Prove that G is a subgroup of SL2 (R).

Solution.
 
cos(θ) − sin(θ)
• Since det sin(θ) cos(θ)
= cos2 (θ) + sin2 (θ) = 1, we get that G ⊆ SL2 (R).
• Setting θ = 0 shows that G contains the identity.
• Since    
cos(θ) − sin(θ) cos(−θ) − sin(−θ)
sin(θ) cos(θ)
· sin(−θ) cos(−θ) = ( 10 01 ),

G is closed under taking inverses.

• We have
     
cos(θ) − sin(θ) cos(ϕ) − sin(ϕ) cos(θ) cos(ϕ)−sin(θ) sin(ϕ) − sin(θ) cos(ϕ)−cos(θ) sin(ϕ)
sin(θ) cos(θ)
· sin(ϕ) cos(ϕ) = sin(θ) cos(ϕ)+cos(θ) sin(ϕ) cos(θ) cos(ϕ)−sin(θ) sin(ϕ)
 
cos(θ+ϕ) − sin(θ+ϕ)
= sin(θ+ϕ) cos(θ+ϕ) .

Thus G is closed under taking products, and thus G is a subgroup of SL2 (R).

3.44. List the subgroups of the quaternion group Q8 .

Solution.
{{1}, {±1}, {±1, ±i}, {±1, ±j}, {±1, ±k}, Q8 } .

3.46. Prove or disprove: if H and K are subgroups of a group G, then H ∪ K is a subgroup of G.

Solution. This is only true if H ⊆ K or K ⊆ H. It suffices to give a counterexample: if G = Z6 ,
H = {0, 2, 4} and K = {0, 3} then H ∪ K = {0, 2, 3, 4} is not a subgroup since it’s not closed under
addition.

3.52. Prove or disprove: every proper subgroup of a nonabelian group is nonabelian.

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 Solution. False. For example, {±1, ±i} ⊂ Q8 is abelian but Q8 is not.

3.54. Let H be a subgroup of G. If g ∈ G, show that gHg −1 = {g −1 hg : h ∈ H} is also a subgroup of G.

Solution.

• Note that gHg −1 is a subset of G since G is closed under multiplication.

• Since 1 ∈ H, we have 1 = g · 1 · g −1 ∈ gHg −1 .
• If ghg −1 , gh0 g −1 ∈ gHg −1 then ghg −1 gh0 g −1 = ghh0 g −1 ∈ gHg −1 since H is closed under multi-
plication.
• If ghg −1 ∈ gHg −1 then (ghg −1 )−1 = gh−1 g −1 ∈ gHg −1 since H is closed under taking inverses.