Name:
Abstract Algebra
Math 481 Winter 2004
Professor Ben Richert
Exam 2
February 27, 2004
Please do all your work in this booklet and show all the steps.
Calculators not allowed.
You may use one side of one half of a standard sheet of note paper.
Problem Possible points Score
1
20
2
15
3
10
4
10
5
10
6
15
Total
80
2
Problem 1. (20 pts.) Consider the group G = S3 ⊕ Z6 .
(a – 5 pts) What is the order of G?
Solution. The order of G is given by the |S3 ||Z6 | = 36. Note that G = {(a, b) |a ∈ S3 , b ∈ Z6 }, and we can easily
count this set.
(b – 5 pts) Is G cyclic?
Solution. Certainly not. The order of any element (a, b) ∈ G is lcm{|a|, |b|}, and as both |a| and |b| divide 6
(the order of both S3 and Z6 ), the order of any element is at most 6. Because the group is order 36, it is not
cyclic.
(c – 5 pts) Let H ≤ S3 be the subgroup of S3 consisting of elements H = {(1), (123), (132)}, and K ≤ Z6 be the subgroup of
Z6 generated by 3, K = h3i. Show that the set H ⊕ K = {(h, k) | h ∈ H, k ∈ K} is a subgroup of G.
Solution. We use the usual criterion. First note that H ⊕ K 6= ∅ because ((1), 0) ∈ H ⊕ K. Suppose then that
(a, b), (c, d) ∈ H⊕K. We note that a, c ∈ H and b, d ∈ K by definition. Then because H and K are subgroups of G,
we know that c−1 ∈ H and −d ∈ K, and that ac−1 ∈ H and b−d ∈ K. Thus (a, b)(c, d)−1 = (ac−1 , b−d) ∈ H ⊕K,
and we conclude that H ⊕ K is a subgroup as required.
!
(d – 5 pts) What is the order of (12), 2 H ⊕ K as an element in G/(H ⊕ K) (it turns out that H ⊕ K is normal in G)?
Solution. The order of this element is the smallest n such that (12)n and 2n are equal to (1) and 0(mod 6) respectively. This is the least common multiple of the order of (12) and 2, which we easily compute: lcm{|(12)|, |2|} =
lcm{2, 3} = 6.
√
Problem 2. (15 pts.) Let G be the group {a + b 3 | a, b ∈ Z} under addition. Show that G is isomorphic to Z ⊕ Z.
√
Solution. Let φ : G → Z ⊕√Z be the map
√ given by a + b 3 → (a, b). We show that this map is an isomorphism.
First we note that if a + b 3 = c + d 3, then it must be that a = c and b = d. Obviously if a = c then b = d and vice versa.
√
√
a−c √
Now because a − c = (b − d) 3, if b 6= d, then it must be the case that
= 3, that is, then 3 is rational. Proving that
b−d
√
m √
= 3. This
3 is not rational isn’t really the point, but if it were then there would be m, n ∈ Z such that (m, n) = 1 and
n
m2
implies that 3 = 2 , so that 3n2 = m2 . Because 3 is prime we know that 3|m, and thus there is t ∈ Z such that 3t = m.
n
Thus 3n2 = 33 t2 , and hence n2 = 3t2 , but this implies that 3|n2 . Again because 3 is prime, we have that 3|n, a contradiction
as (n, m) = 1.
√
√
Once we have this fact, the rest of the proof √
follows easily. The map is well √
defined because if a + b 3 = c + d 3, then a = c
and b =√d (as stated above)
so that φ(a + b 3) = (a, b) = (c, d) = φ(c + d 3) as
√
√ required. √The map is injective because if
φ(a + b 3) = φ(c + d 3), then (a, b) = (c, d), so a = c and
b
=
d,
that
is,
a
+
b
3 = c + d 3. The map is obviously onto.
√
√
√
√ √
Finally, if a + b 3 and c + d 3 are elements of G, then φ (a + b 3) + (c + d 3) = φ (a + c) + (b + d) 3 = (a + c, b + d) =
√
√
(a, b) + (c, d) = φ(a + b 3) + φ(c + d 3).
Problem 3. (10 pts.) Is Z10 = h2i × h5i?
Solution. We know that Z10 = h2i × h5i if h2i + h5i = Z10 and h2i ∩ h5i = {0}. The latter fact is obvious when we write
out h2i = {0, 2, 4, 6, 8} and h5i = {0, 5}. To see that h2i + h5i = Z10 its probably fastest to write out the sum as a set:
h2i + h5i = {0 + 0 = 0, 2 + 0 = 2, 4 + 0 = 4, 6 + 0 = 6, 8 + 0 = 8, 0 + 5 = 5, 2 + 5 = 7, 4 + 5 = 9, 6 + 5 = 1, 8 + 5 = 3}. This is
obviously all of G.
Problem 4. (10 pts.) Suppose that G is a group of order 324, a ∈ G, hai is normal in G, and |G/hai| = 18.
(a – 5 pts.) How many left cosets of ha9 i are there?
3
G
|G|
324
18
18
9
Solution. We know that =
= 18, so |a| =
= 18. Thus the order of a =
=
= 2 (we
hai |a|
18
(9, 18)
9
|c|
are using an old theorem about cyclic groups here, that |cd | =
). The number of cosets of ha9 i is thus
(|c|,
d)
G
|G|
324
hai = |ha9 i| = 2 = 162.
(b – 5 pts.) If H and K are proper subgroups of G, H ≤ K, and |H| = 2, what are the possible orders of K?
Solution. We can factor |G| = 324 as 34 22 , and we know that |K| divides |G| and |H| = 2 divides |K|. So the
possible orders of K are 2i 3j where 1 ≤ i ≤ 2, 0 ≤ j ≤ 4, with the exception of the case i = 2, j = 4 (as then K
would not be proper).
Problem 5. (10 pts.) Let G be a group and let G0 be the group generated by the elements {x−1 y −1 xy | x, y ∈ G}. It can be
shown that G0 is normal in G. Prove that G/G0 is Abelian.
Solution. We are required to show that for all h1 G0 , h2 G0 ∈ G/G0 , we have that h1 G0 h2 G0 = h2 G0 h1 G0 . This happens if and
−1
0
0
0
only if h1 h2 G0 = h2 h1 G0 , that is, if h1 h2 (h2 h1 )−1 = h1 h2 ‘h−1
1 h2 ∈ G . But this is true by the definition of G , so G/G is
Abelian.
Problem 6. (15 pts.) Let φ : G → G0 be an isomorphism of groups.
n
o
φ(g1 ), . . . , φ(gn ) generates G0 .
n
o
Solution. It is enough to show that if {g1 , . . . , gn } generates G then φ(g1 ), . . . , φ(gn ) generates G0 , this because
(a – 10 pts.) Prove that the set {g1 , . . . , gn } generates G if and only if the set
φ−1 is also an isomorphism. So suppose that {g1 , . . . , gn } generates G and let g 0 ∈ G0 . Then we must demonstrate
that there are integers a1 , . . . , an such that φ(g1 )a1 · · · φ(gn )an = g 0 . Note that because φ is onto there is g ∈ G
such that φ(g) = g 0 , and because {g1 , . . . , gn } generates G there are integers a1 , . . . , an such that g1a1 · · · gnan = g.
an
Then φ(g1a1 ) · · · φ(gnan ) = φ(g), that is,nφ(g1 )a1 · · · φ(gn )o
= g 0 (we use here that φ is an isomorphism, so that
φ(giai ) = φ(gi )ai ) and we conclude that φ(g1 ), . . . , φ(gn ) generates G0 as required.
(b – 5 pts.) Use the fact established in part (a) to prove that Z4 ∼
6 Z2 ⊕ Z2 .
=
Solution. We note that Z4 = h1i is cyclic and hence is generated by a set of size one. On the other hand, a simple
check demonstrates that Z2 ⊕ Z2 is not cyclic, and thus is not generated by a single element. Of course, no function
can take a set of size one to a set of size two, so we conclude that no isomorphism exists between Z4 → Z2 ⊕ Z2 .