Download Math 305, Fall 2007 Midterm 2 Review Solutions

Math 305, Fall 2007
Midterm 2 Review Solutions
(1) (a) Disprove: For all finite groups G and all m ∈ Z≥1 , if m||G|, then there is a ∈ G such that o(a) = m.
(Hint: Look at the quaternions Q8 .)
(b) Prove that the statement in (b) is true if G = Zn for any n ∈ Z≥2 .
Solution:
(a) Let G = Q8 and m = 8. Then clearly G is not cyclic, and hence for all a ∈ G we have o(a) < 8.
(b) First claim is that, in a finite cyclic group hai of order n, if k ∈ Z, then hak i is a subgroup of order
n/(n, k) (think about this). Hence for the cyclic group Zn , since [1] is a generator, we know that [k]
generates a subgroup of order n/(n, k) for all k ∈ Z. Let m|n. Then d = n/m certainly divides n,
so (n, d) = d. By the above, the element [d] generates a subgroup of order n/(n, d) = n/d = m, as
required.
(2) Construct the subgroup lattices (see page 90 in Durbin) for Z2 × Z4 and for A4 .
Solution: For Z2 × Z4 , we have the lattice
Z2 × Z4 R
ppp
ppp
p
p
ppp
h([1], [1])i
h([0], [1])i
RRR
RRR
RRR
RRR
h([1], [2]), ([1], [0])i
qq
ggggg mmm
qqq
ggggg mmmmm
g
q
g
g
q
g
q
gg
mm
mmm
qqqggggggg
h([1], [2])i
h([0], [2])i
h([1], [0])i
MMM
m
m
m
MMM
m
MMM
mmm
mmm
MM
m
m
m
h([0], [0])i
The lattice for A4 is as
m A48QKKQQQ
88 KK QQQ
mmm
m
m
88 KKK QQQQ
mm
m
m
m
88 KKK QQQQ
mmm
Q
KK
88
KK QQQQQ
88
h(12)(34), (13)(24)i
KK
QQQ
KK
88
QQQ
KK
nnnzzz
QQQ
n
8
n
K
n
8
z
QQQ
K
n
z
K
8
n
QQQ
n
K
z
8
K
z
nn
n
z
n
z
n
n
z
h(134)i
h(234)i
h(123)i
h(124)i
nnn
zz
nnn
zz
n
mm
tt
z
n
m
n
m
t
z
m
n
t
z
n
mm
tt
zz
nnn
mmm
tt
m
m
t
nnn
zz
m
t
mmm
tt
h(12)(34)i ZZZZZZh(13)(24)i XXX
h(14)(23)i
tt mmmmm
t
ZZZZZZZ
Q
X
t
QQQ
tt mmmm
ZZZZZZZXXXXXXXXX
QQ
ZZZZZZZ XXXXX
tttmtmmmm
ZZZZZZZXXXXX QQQQQ
t m
ZZZZZZXZXXXX QQ
tmtmmm
ZZZZXZXZX
h(1)i
(3) Recall that a nontrivial subgroup H of G is proper if H 6= G. Construct a proper subgroup of Q that is
not cyclic. Hint: consider the dyadic rationals, fractions whose denominators are powers of 2.
Solution: Consider the subset H of Q given by {m/2n : m ∈ Z, n ∈ Z≥0 }. Certainly this subset is nonempty.
Consider a = m/2n and b = r/2s , where m, r ∈ Z and r, s ∈ Z≥0 . Then a − b = (2s m − 2n r)/2n+s , so a − b
belongs to H. Hence H is a subgroup. Suppose that H = hm/2n i, where m ∈ Z is odd and n ∈ Z≥0 . Then
clearly m/2n+1 does not belong to hm/2n i but does belong to H, a contradiction.
(4) 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.
(5) Compute the cosets of H in G and the index [G : H] for the following groups G and subgroups H.
(a) G = S4 and H = S3
(b) G = Z6 × Z4 and H = h[2]i × h[2]i
(c) G = Z × Z2 and H = h(1, [1])i
(d) G = R × Z and H = R × {0}
Solution:
(a) Set of
(b) Set of
(c) Set of
(d) Set of
(e) Set of
cosets
cosets
cosets
cosets
cosets
is
is
is
is
is
{H, H(14), H(24), H(34)} and [G : H] = 4.
{H, Hρ, Hρ2 , Hρ3 , Hρ4 } and [G : H] = 5.
{H, H + ([0], [1]), H + ([1], [0]), H + ([1], [1])} and [G : H] = 4.
{H, H + (1, [0])} and [G : H] = 2.
{. . . , H + (0, −1), H, H + (0, 1), H + (0, 2), . . .} and [G : H] = ∞.
(6) (a) Consider φ : Z6 → Z3 given by φ([a]6 ) = [a]3 . Prove that φ is a well-defined homomorphism and
compute ker φ.
(b) Prove that φ : Z3 → Z6 given by φ([a]3 ) = [a]6 is not well-defined.
(c) 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) (a) Let the dihedral group of 2n elements, denoted by Dn , be 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 turns out to be the symmetry group of a regular n-gon (so D4 is the group of symmetries
of a square. Prove that A4 and D12 are not isomorphic. (Hint: Look at elements of order 2.)
(b) Define SL2 (Z3 ) to be the set of 2 × 2 matrices A with entries in Z3 such that det A = 1 (as computed
in Z3 ); i.e. add and multiply mod 3). Prove that SL2 (Z3 ) is a group that is not isomorphic to S4 .
You may assume that SL2 (Z3 ) has 24 elements. (Hint: Look at elements of order 6.)
Solution:
(a) The only elements of order 2 in A4 are (12)(34), (13)(24) and (14)(23). But there are seven elements
of order 2 in D12 , namely σ, ρσ, ρ2 σ, ρ3 σ, ρ4 σ, ρ5 σ and ρ3 . •
(b) The group S4 has no element of order 6. However the matrix
−1
0
1 −1
has order 6 in SL2 (Z3 ).
(8) Prove that the following pairs of groups are not isomorphic.
(a) D8 and Q8
(b) R and Z
(c) D12 and S4
(d) Z × Z2 and Z × Z3
Solution:
(a) The group Q8 has only one element of order 2, while D8 has five elements of order 2.
(b) The group Z is cyclic but R is not.
(c) In D24 the element ρ has order 12, but in S4 there are no elements of order 12.
(d) The group Z × Z2 has an element of order 2, namely (0, [1]). However, no element of Z × Z3 has order
2.
(9) 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 .
(10) Determine (with proof) whether the following are automorphisms.
(a) Define φ : S3 → S3 by φ(α) = α−1 .
(b) Define φ : Z12 → Z12 by φ([a]) = [3a].
Solution:
(a) Let x = (12) and y = (13). Then φ(xy) = φ((132)) = (123) and φ(x)φ(y) = (12)(13) = (132), so φ is
not a homomorphism and thus not an automorphism.
(b) Notice that φ([0]) = φ([4]) = 0, but [0] 6= [4] in Z12 , so φ is not injective. Hence it is not an
automorphism.
(11) Let G be a group and let H and K be normal subgroups with H ∩ K = {e}. Prove that xy = yx for all
x ∈ H and y ∈ K. (Hint: Show that x−1 y −1 xy ∈ H ∩ K.)
Solution: Let x ∈ H and y ∈ K. Then x−1 y −1 x lies in K, so x−1 y −1 xy ∈ K. Also y −1 xy ∈ H, so
x−1 y −1 xy ∈ H. Therefore x−1 y −1 xy ∈ H ∩ K. So x−1 y −1 xy = e. Hence xy = yx.
(12) Let G be a group all of whose subgroups generated by a single element are normal. For a, b ∈ G, prove
that there is k ∈ Z such that ab = bak .
Solution: Let a, b ∈ G. Then hai is normal in G. Hence b−1 ab ∈ hai, so b−1 ab = ak for some k ∈ Z.
Therefore we have ab = bak .
(13) Let H be the subset of M (R) consisting of continuous functions f : R → R. Decide if H is normal in
(M (R), +).
Solution: Since (M (R), +) is abelian, all subgroups are normal.
1 0
(14) Find the order of 12Z + 8 in Z/12Z and the order of SL2 (R)
in GL2 (R)/SL2 (R).
0 2
Solution: 3 and ∞.
(15) 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.
(16) Let G be a group and recall that Z(G) is the center of G. Also recall from class 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.
(17) Let A ⊆ G be nonempty and define NG (A) = {g ∈ G : gAg −1 = A}. Here gAg −1 = {gag −1 : a ∈ A}. We
call NG (A) the normalizer of A in G.
(a) Prove that, for any nonempty A ⊆ G, the set NG (A) is a subgroup. Note that A does not have to be
a subgroup.
(b) Prove that, if H is a subgroup of G, then it is a subgroup of NG (H).
(c) Let CG (H) be the set of elements of G which commute with all elements of H. Prove that, if H is a
subgroup, then CG (H) is a subgroup of NG (H), but that equality does not always hold.
(d) Prove that, if H is a normal subgroup of G, then NG (H) = G.
(e) Find the normalizer of each subgroup of S3 and Q8 .
Solution:
(a) Let g, h ∈ NG (A). Then gAg −1 = A and hAh−1 = A. Note that h−1 Ah = A as well. So h−1 ∈ NG (A).
Also ghA(gh)−1 ghAh−1 g −1 = gAg −1 = A, so gh ∈ NG (A), so NG (A) is closed. Certainly eAe−1 = A,
so NG (A) has the identity. Therefore it is a subgroup of G.
(b) It suffices to show that H ⊆ NG (H). Let h ∈ H. Then hH = H = Hh, so hHh−1 = H, so h ∈ NG (H).
(c) Let g ∈ CG (H). Then g commutes with every element h ∈ H. In particular, we have gH = Hg, so
gHg −1 = H. Therefore g ∈ NG (H). An example in which equality does not hold is H = A3 and
G = S3 .
(d) Let H be normal in G. Let g ∈ G. Then gH = Hg, so gHg −1 = H. Therefore g ∈ NG (H). So
G = NG (H).
(e) In S3 , the normalizer of {e}, h(123)i and S3 is S3 itself. Each of the other subgroups is its own
normalizer. In Q8 , the normalizer of every subgroup is Q8 .
(18) Let (G, ◦) be the group of all functions τa,b : R → R given by τa,b (x) = ax + b, where b ∈ R and a ∈ R× .
Let N be the subset N = {τ1,b ∈ G : b ∈ R}. Prove that N C G and that G/N ∼
= R× .
Solution: Let φ : G → R× be given by φ(τa,b ) = a for all τa,b ∈ G. It is not hard to show that this map
is a well-defined homomorphism with kernel N . Also the image is R× . Hence by the First Isomorphism
Theorem, we have G/N ∼
= R× .