Math 30710
Practice Exam 1 Solutions
October 2, 2013
Name
This is a 50-minute exam. Books and notes are not allowed. Make sure that your work is legible, and
make sure that it is clearly marked where your answers are. Show all work. If something is not clear, ASK
ME!! Good luck!
1. For each of the following relations, decide if it is an equivalence relation or not. If it is, you don’t
have to prove it, but describe the equivalence classes. If it is not, explain why not. You are allowed
to quote theorems from class.
(a) xRy in R if x4 = y 4 .
Yes. The equivalence classes are all pairs of numbers {x, −x} where x ∈ R.
(b) xRy in Z if x + y is divisible by 3.
No. It is not reflexive or transitive in general. For example, 2R2 is not true, and 1R2 and 2R4
are true, but 1R4 is not true.
(c) xRy in Z if x − y is divisible by 3.
Yes. The equivalence classes are the congruence classes mod 3.
(d) If G is a group and H is a subgroup, and if a, b ∈ G, then aRb if a−1 b ∈ H.
Yes. In class we saw that this defines an equivalence relation, and that the equivalence classes
are the cosets aH.
2. Above is the group table for the dihedral group D4 . For each of the following, either give an example
to show that it exists, or give a simple (but solid) reason why it does not exist.
1
2
(a) A subgroup isomorphic to Z2 .
Yes. For example, hµ1 i.
(b) A subgroup isomorphic to Z4 .
Yes. For example, hρ1 i.
(c) A subgroup isomorphic to Z8 .
No. By inspecting the multiplication table we see that there is no element of order 8.
(d) A subgroup isomorphic to S3 .
No. By Lagrange’s theorem, the order of a subgroup divides the order of the group, and 6 does
not divide 8.
3. Consider the permutation
σ=
1 2 3 4 5 6 7 8 9 10
8 7 4 2 10 1 9 6 5 3
(a) Find σ −1 and write it in the space below:
σ −1 =
1 2 3 4 5 6 7 8 9 10
6 4 10 3 9 8 2 1 7 5
(b) Write σ as a product of disjoint cycles. (Make sure you look at σ and not σ −1 .)
σ = (1, 8, 6)(2, 7, 9, 5, 10, 3, 4, ).
(c) Write σ as a product of transpositions.
σ = (1, 6)(1, 8)(2, 4)(2, 3)(2, 10)(2, 5)(2, 9)(2, 7).
(d) Find the order of σ.
21.
4. Let G be a group. Prove that if G has only finitely many subgroups then G must be a finite group.
(Hint: think about why Z has infinitely many subgroups.)
Suppose that G were infinite. Let a ∈ G and consider the subgroup hai. If this is infinite then
it is isomorphic to Z, which has infinitely many subgroups (2Z, 3Z, 4Z, ...); hence G has infinitely
many subgroups, contradicting the hypothesis. So G has no infinite cyclic subgroup. But now every
element of G generates a cyclic subgroup (these subgroups are not necessarily distinct), so G is the
union of all its cyclic subgroups. Since there are only finitely many of them, and since each of them
is finite, we get that G must be finite.
3
5. Prove that if G is a finite group of order n with identity element e then an = e for all a ∈ G.
Let a ∈ G and consider the subgroup hai. Let m = |hai|. We know that am = e because for a
finite cyclic group, the order of a generator is equal to the order of the group. We also know that m
divides n by Lagrange’s theorem. Thus n = mk for some positive integer k, so
an = amk = (am )k = ek = e.
6. Let G = Z20 .
(a) List all the subgroups of G.
{0}
Z20 = {0, 1, 2, . . . , 19} = h1i = h3i = h7i = h9i = h11i = h13i = h17i = h19i
{0, 2, 4, 6, 8, 10, 12, 14, 16, 18} = h2 = h6i = h14i = h18i
{0, 4, 8, 12, 16} = h4i = h8i = h12i = h16i
{0, 5, 10, 15} = h5i = h15i
{0, 10} = h10i
(b) What theorem tells you that the subgroups you listed are all the possible subgroups? (Hint:
why isn’t there a subgroup isomorphic to the Klein 4-group?)
Every subgroup of a cyclic group is cyclic.
(c) Draw the subgroup diagram for Z20 . Make sure all of the subgroups listed in (a) appear in this
diagram!
h1i
h2i
h4i
h5i
h10i
h0i