Name: Exam I
Instructions. Answer each of the questions on your own paper and put your name on
each page of your paper.
1. Give an example of each of the following. No proofs are required for this exercise only.
(a) A nonabelian group of order 12.
I Solution. Two possible examples are the dihedral group D12 of order 12, and
the alternating group A4 . J
(b) An abelian, but noncyclic group of order 12.
I Solution. Z2 × Z6 . Several of you gave Z3 × Z4 as an example. But this group
is in fact cyclic of order 12. (See Exercise 11 (c).) J
(c) A normal subgroup H of the symmetric group S3 , and a non-normal subgroup K
of the symmetric group S3 .
I Solution. H = h(1 2 3)i and K = h(1 2)i. J
(d) An element σ of order 12 in the alternating group A10 .
I Solution. σ = (1 2 3 4)(5 6 7 8 9 10) or σ = (1 2 3 4)(5 6 7)(8 9) J
(e) An element b of order 5 in the cyclic group G = hai of order 20.
I Solution. b = a4 . J
2. Let G be the group of invertible 2 × 2 upper triangular matrices with entries in R.
That is, ½· ¸ ¾
a b
G= : a, b, d ∈ R, with ad 6= 0 .
0 d
Let ½· ¸ ¾
a 0
D= : ad 6= 0 ⊆ G
0 d
be the subgroup of invertible
· ¸ diagonal matrices and let U ⊆ G be the subgroup of
1 b
matrices of the form where b ∈ R is arbitrary. For this exercise, you may
0 1
assume without proof that G, D, and U are groups under the operation of matrix
multiplication.
µ· ¸¶ · ¸
a b a 0
(a) Define a function ϕ : G → G by ϕ = . Show that ϕ is a group
0 d 0 d
homomorphism.
Math 7200 September 29, 2006 1
Name: Exam I
µ· ¸· ¸¶ µ· ¸¶ · ¸
a1 b1 a2 b2 a1 a2 a1 b2 + b1 d2 a1 a2 0
I Solution. ϕ =ϕ = =
· ¸· ¸ 0 µ·
d1 0 d¸¶ 2
µ· 0
¸¶ d1 d2 0 d1 d2
a1 0 a2 0 a 1 b1 a2 b2
=ϕ ϕ . Thus ϕ is a group homomor-
0 d1 0 d2 0 d1 0 d2
phism. J
(b) Show that U is a normal subgroup of G and that G/U is isomorphic to D.
· ¸
a b
I Solution. Since ∈ Ker(ϕ) if and only if a = d = 1, it follows that U =
0 d
Ker(ϕ). Since kernels of homomorphisms are normal subgroups, it follows that
U is a normal subgroup of G. Moreover, D = Im(ϕ), so by the first isomorphism
theorem,
G/U = G/ Ker(ϕ) ∼ = Im(ϕ) = D.
J
(c) True or False (with justification): G is the internal direct product of U and D.
I Solution. The criterion for G to be the internal direct product of U and D
(Definition
· ¸ 6.3) requires that
· both
¸ U and D are normal subgroups of G. But
1 0 1 1
A= ∈ D, and B = ∈ G, but
0 2 0 1
· ¸· ¸· ¸ · ¸
−1 1 1 1 0 1 −1 1 1
BAB = = ∈
/ D,
0 1 0 2 0 1 0 2
so D is not normal in G. Therefore, G is not the internal direct product of U
and D. J
3. Let G be a group of order 2006 = 2 × 17 × 59. Can a group of this order be simple?
Prove that your answer is correct. (Recall that a group G is simple if the only normal
subgroups of G are {e} and G.)
I Solution. Let n59 (G) be the number of 59-Sylow subgroups of G. According the
Sylow’s theorem, n59 (G) ≡ 1 (mod 59) and n59 (G)|34 = 2006/59. Thus, n59 (G) = 1
so there is exactly one subgroup of G of order 59. Let H be this subgroup of order 59.
Since conjugating a subgroup of order 59 produces another subgroup of order 59, it
follows that gHg −1 = H for all g ∈ G, so H is normal in G. Thus G is not simple. J
4. (a) Let G be a group and a ∈ G. State (without proof) the formula that relates the
size of the conjugacy class [a]C of a and the size of the centralizer C(a) of a.
I Solution. |[a]C | = [G : C(a)] = |G| / |C(a)|, assuming that |G| < ∞ (Page
15). J
(b) If σ = (1 2)(3 4) ∈ S4 , compute the number of elements of S4 that are conjugate
to σ and the number of elements of S4 that commute with σ.
Math 7200 September 29, 2006 2
Name: Exam I
I Solution. The permutations that are conjugate to σ are all of those with the
same cyclic type. Since σ is a product of two disjoint 2-cycles in S4 , there are
µ ¶
1 4·3 2·1
× =3
2 2 2
elements in the conjugacy class [σ]C . The elements of S4 that commute with σ
are the members of the centralizer C(σ), and by the formula in part (a),
|S4 | 24
|C(σ)| = = = 8.
|[σ]C | 3
It was not part of the question, but it is easy to list the 8 permutations that
commute with σ: (1), (1 2)(3 4), (1 2), (3 4), (1 3)(2 4), (1 4)(2 3), (1 3 2 4),
(1 4 2 3). To see where these come from, note that the first, second, third, and
fourth elements form a subgroup of S4 of order 4, and hence all elements commute.
The same is true of the first, second, fifth and sixth elements, and the first, second,
seventh and eighth elements. J
Math 7200 September 29, 2006 3
