Download Math 311 Final Problem Set – Solution December 2002

Math 311
Final Problem Set – Solution
December 2002
Instructions: You may refer to your class notes and to Joseph Gallian’s Contemporary
Abstract Algebra in completing this problem set. You may not refer to any other books or
notes, and you may not get help from other people.
Part I: True or false
(5 points each; no justification required)
1. Every subgroup of an Abelian group is Abelian
True.
2. If G is a group and d divides the order of G, then G contains an element of order d.
False. The group S3 has order 6, but contains no element of order 6.
3. For each n ≥ 3, the center of Sn is trivial.
True. An element σ of Sn is conjugate to every other element with the same cycle
structure. If σ ∈ Z(Sn ), then σ can be conjugate only to itself. The only element in
Sn (n ≥ 3) whose cycle structure is not shared with any other element is the identity.
4. Let G be a group and H be a subgroup of G. Let a and b be elements of G. If
aH = bH, then Ha = Hb.
False. Take, for example H = {e, (1 2)} in S3 . Then
(1 3)H = (1 2 3)H = {(1 3), (1 2 3)}
while
H(1 3) = {(1 3), (1 3 2)}
and
H(1 2 3) = {(2 3), (1 2 3)}
so that H(1 2 3) 6= H(1 3).
5. The groups U (20) and U (24) are isomorphic.
False. In U (24), every non-identity element has order 2, while in U (20), the element 3
has order 4.
Part II: Computation
(10 points each)
1. The quaternion group Q consists of these eight elements: 1, −1, i, −i, j, −j, k, and
−k. The operation is multiplication, and products among i, j, and k are computed
using the rules
i2 = j 2 = k 2 = −1
ij = k, jk = i, ki = j
ji = −k, kj = −i, ik = −j.
(Multiplication by 1 and −1 is just ordinary multiplication.)
Draw a Cayley digraph for Q on the generating set {i, k}.
Solution: Here is the Cayley graph. The red edges represent i and the green edges
represent k.
-j
-k
k
-1
i
-i
1
j
2. List all the elements of order 20 in Z6200 .
Solution: The elements of order 20 in Z6200 are the generators of the subgroup h310i,
which is the unique subgroup of order 20 in Z6200 . One generator of this subgroup is
6200/20 = 310.
The other generators are of the form 310n, where n is an integer in {0, 1, 2, 3, . . . , 19}
that is relatively prime to 20. The complete list of elements of order 20 in Z6200 is
310 × 1
310 × 3
310 × 7
310 × 9
310 × 11
310 × 13
310 × 17
310 × 19
=
=
=
=
=
=
=
=
310
930
2170
2790
3410
4030
5270
5890
3. Find |cl((1 2 3)(4 5 6 7))| in S10 .
Solution: We recall that in Sn , any two elements with the same cycle structure are
conjugate. Thus the conjugacy class of (1 2 3)(4 5 6 7) comprises all elements of S10 of
the whose cycle structure is (· · ·)(· · · ·).
The number of such elements is
10 × 9 × 8 7 × 6 × 5 × 4
×
= 50, 400.
3
4
Part III: Proofs
(10 points each; presentation counts. You may cite theorems proved in the textbook to
support your proofs.)
1. Let G be a group with identity element e. Let a, b and c be elements of G such that
abc = e. Prove that bca = e.
Proof: We have a(bc) = e, which implies that bc = a−1 . Since a left inverse is also a
right inverse, this implies that (bc)a = e, which is what we were asked to prove.
2. Let G1 and G2 be groups and let ϕ be a map from G1 to G2 . Suppose ϕ is operationpreserving and onto. Let Z(G1 ) denote the center of G1 and Z(G2 ) denote the center
of G2 . Let H = {ϕ(a) : a ∈ Z(G1 )}. Prove that H ⊆ Z(G2 ).
Proof: Let h ∈ H. Then h = ϕ(x) for some x ∈ Z(G1 ).
We need to show that h ∈ Z(G2 ). That is, we need to show that h commutes with
every element of G2 . Let g ∈ G2 . Since ϕ is onto, we know that g = ϕ(y) for some
y ∈ G1 .
Since x ∈ Z(G1 ) and y ∈ G1 , we know that xy = yx.
Now we use the fact that ϕ is operation-preserving to get
gh =
=
=
=
=
ϕ(y)ϕ(x)
ϕ(yx)
ϕ(xy)
ϕ(x)ϕ(y)
hg.
Since h commutes with an arbitrary element of G2 , we conclude that h ∈ Z(G2 ), and
therefore that H ⊆ Z(G2 ).
3. Let G be a group of permutations of the set {1, 2, 3, . . . , 20}. Let g be some element
of G such that g(1) = 3. Let H = stabG (1). Prove that gH = {x ∈ G : x(1) = 3}.
Proof: Let S = {x ∈ G : x(1) = 3}. We show first that gH ⊆ S.
Let y ∈ gH. Then y = gh for some h ∈ H. Using the fact that h fixes the letter 1, we
get
y(1) = gh(1)
= g(h(1))
= g(1)
= 3.
Thus y ∈ S, and we have shown that gH ⊆ S.
We now show that S ⊆ gH. Let y ∈ S. Then we know that y(1) = 3. We also know
that g −1 (3) = 1, so we can write
g −1 y(1) = g −1 (3)
= 1.
That is, the element g −1 y fixes the letter 1, so that g −1 y ∈ H. By the definitions of
cosets, then,
g(g −1 y) ∈ gH.
But gg −1 y = y. So we have shown that y ∈ gH. That is, S ⊆ gH.
Having shown containment in both directions, we conclude that S = gH.
4. A loop in a Schreier diagram is a directed edge that leads from some coset back to the
same coset.
Suppose G is a group, N is a normal subgroup of G, and S = {s1 , s2 , . . . , sr } is a set
of elements of G such that S ∩ N is empty. Prove that the Schreier diagram for G
modulo N on the set S has no loops.
Proof: Suppose there is a loop in the Schreier diagram for G modulo N on the set S.
Then there exists some g ∈ G and some s ∈ S such that N gs = N g.
In particular, this implies that gs ∈ N g. Now we know that N is normal, so N g = gN ,
and so we have gs ∈ gN . That is, there is some n ∈ N such that gs = gn. By left
cancellation, then, s = n.
This implies that s ∈ N , contradicting the hypothesis that S ∩ N is empty.
Thus no loop can exist in this Schreier diagram.
5. Let G be a group and Z(G) be the center of G. Prove that |G : Z(G)| is not a prime.
Proof: Suppose |G : Z(G)| = p, where p is a prime. Then every element of G/Z(G)
has order 1 or p, so that G/Z(G) is cyclic.
From Theorem 9.3 in Gallian, it follows that G is Abelian, so that Z(G) is actually
equal to G. This means that |G : Z(G)| = 1, which contradicts the hypothesis that
|G : Z(G)| is prime.