Math 306 Topics in Algebra, Spring 2013
Homework 1 Solutions
(1) (4 pts) Compute the number of elements in Z100 /h[15]i.
Solution: Since the order of [15] in Z100 is 20, the order of the quotient group is |Z100 |/|h[15]i| = 100/20 = 5.
(2) (4 pts) Compute the cosets of S3 in S4 and find the index [S4 : S3 ]
Solution: Set of cosets is {S3 , S3 (14), S3 (24), S3 (34)} and [S4 : S3 ] = 4.
(3) (4 pts/part)
(a) Show that Sn is nonabelian for n ≥ 3.
(b) Decide whether (245)(1354)(125) ∈ S5 is an even or an odd permutation.
Solution:
(a) Cycles (12) and (13) do not commute since (12)(13) = (132) and (13)(12) = (123).
(b) Since (245) = (25)(24), (1354) = (14)(15)(13), and (125) = (15)(12), the total number of transpositions
is 7, so the permutation is odd. (Alternatively, (245)(1354)(125) = (14)(253) = (14)(53)(23).)
(4) (4 pts/part)
(a) For G a group with operation ∗ and for a ∈ G define the centralizer of a in G to be the set
C(a) = {x ∈ g : a ∗ x = x ∗ a}.
Show that C(a) is a subgroup of G.
(b) Find the centralizer of 2 in Z.
Solution:
(a) Since a ∗ e = e ∗ a, e ∈ C(a) and thus C(a) 6= ∅. If x, y ∈ C(a), then
a ∗ (x ∗ y) = (a ∗ x) ∗ y = (x ∗ a) ∗ y = x ∗ (a ∗ y) = x ∗ (y ∗ a) = (x ∗ y) ∗ a
and so x ∗ y ∈ C(a). Finally, if x ∈ C(a), then a ∗ x = x ∗ a, so
x−1 ∗ a = x−1 ∗ a ∗ (x ∗ x−1 ) = x−1 ∗ (a ∗ x) ∗ x−1 = x−1 ∗ (x ∗ a) ∗ x−1 = (x−1 ∗ x) ∗ a ∗ x−1 = a ∗ x−1
and so x−1 ∈ C(a). Thus C(a) is a subgroup.
(b) Since Z is abelian, every element commutes with every other element. In particular, everything commutes
with 2, so C(2) = Z.
(5) (5 pts) Let G be a group. Prove that, if G has at least two elements of order 2, then G is not cyclic.
Solution: Suppose that G is cyclic. If G is infinite, then G ∼
= Z, so G has no elements of order 2, a contradiction.
If G is finite, then G ∼
= Zn for some n ∈ Z≥2 . If [a]n ∈ G has order two, then [2a]n = 0. If n is odd, then the
only [a]n satisfying this equation is [a]n = [0]n which has order 1. If n is even, then the only [a]n satisfying this
equation are [a]n = [0]n , which has order 1, or [a]n = [n/2]n , but then there is only one.
(6) (a) (4 pts) Consider φ : Z6 → Z3 given by φ([a]6 ) = [a]3 . Prove that φ is a well-defined homomorphism and
compute ker φ.
(b) (4 pts) Prove that φ : Z3 → Z6 given by φ([a]3 ) = [a]6 is not well-defined.
(c) (5 pts) Let n, m ∈ Z>1 . Determine a necessary and sufficient condition required for the function φ : Zn →
Zm given by φ([a]n ) = [a]m to be well-defined and prove your claim. Your claim should be an iff statement.
Solution:
(a) Let [a]6 = [b]6 in Z6 . Then 6|(a − b). Hence 3|(a − b). so [a]3 = [b]3 in Z3 , so φ([a]6 ) = φ([b]6 ). Hence
φ is well-defined. The kernel of φ is {[0]6 , [3]6 }.
(b) Consider [0]3 and [3]3 . We have [0]3 = [3]3 in Z3 but [0]6 6= [3]6 in Z6 . Hence φ([0]3 ) 6= φ([3]3 ), so φ is
not well-defined.
(c) We claim that φ : Zn → Zm given by φ([a]n ) = [a]m is well-defined iff m|n. First suppose that m|n. If
[a]n = [b]n in Zn , then n|(a − b). Since m|n, we then have m|(a − b), so [a]m = [b]m in Zm . Hence
φ([a]n ) = φ([b]n ). Therefore φ is well-defined. Now suppose that m does not divide n. Consider [n]n and
[0]n in Zn . Clearly [n]n = [0]n in Zn . However, if [n]m = [0]m in Zm , then m|n − 0, a contradiction.
Hence [n]m 6= [0]m in Zm so φ is not well-defined.
(7) (5 pts) Recall that the dihedral group of 2n elements, denoted by Dn , is the group generated by two elements
ρ and σ satisfying the relations ρn = σ 2 = e and σρ = ρ−1 σ. Thus the 2n elements of Dn are
Dn = {e, ρ, ρ2 , ..., ρn−1 , ρσ, ρ2 σ, ..., ρn−1 σ}.
This group is the symmetry group of a regular n-gon (so D4 is the group of symmetries of a square). Prove
that A4 and D6 are not isomorphic. (Hint: Look at elements of order 6.)
Solution: There is one element of order 6 in D6 , namely ρ, and there are no elements of order 6 in A4 .
(8) (5 pts) List all the (isomorphism class representatives of) abelian groups of order 240.
Solution: Note that 240 = 24 · 3 · 5. The abelian groups of order 240 are Z24 × Z3 × Z5 , Z2 × Z23 × Z3 × Z5 ,
Z2 × Z2 × Z22 × Z3 × Z5 , Z22 × Z22 × Z3 × Z5 , and Z2 × Z2 × Z2 × Z2 × Z3 × Z5 .
(9) (5 pts) Let G be a group and recall that Z(G) is the center of G (look up the definition if you don’t remember).
Also recall that Z(G) is a normal subgroup of G, so that we may consider the quotient G/Z(G). Prove that, if
G/Z(G) is cyclic, then G is abelian. (Hint: If G/Z(G) is cyclic with generator Z(G)x, show that every element
of G can be written in the form xn z for some n ∈ Z and z ∈ Z(G).)
Solution: Let g ∈ G and consider Z(G)g ∈ G/Z(G). Since G/Z(G) is cyclic, it is generated by some Z(G)x,
so there is k ∈ Z such that Z(G)g = (Z(G)x)k = Z(G)xk . So g = z1 xk for some z1 ∈ Z(G). Similarly if
h ∈ G, then h = z2 xj for some z2 ∈ Z(G) and j ∈ Z. Therefore gh = z1 xk z2 xj = z1 z2 xj+k = z2 xk z1 xj = hg,
so G is abelian.