Instructions: Do any four of the seven problems. Don’t
forget to put your name on your answer sheets.
[1] Let G be a cyclic group. Let S be the union of all of the proper sub groups
of G .
(a) If G is infinite, show that |G − S| = 2 (i.e., show that there are only two
elements in G that are not in S).
(b) If G is finite, show that |G − S| = Ø(|G|) (i.e., show that there are
Ø(|G|)
elements in G that are not in S).
(c) Prove that a cyclic group G is never a union of proper subgroups.
Answer: (a) If G =< g > is an infinite cyclic group, then it has only two
generators: g, which is given, and g-1. (Clearly,
any subgroup that has either g or g-1 has the other, so one generates if and
only if the other does. But if n is an integer but
not either 1 or −1, then < gn > can’t contain g, since then
implies
that
, and hence that g, and so G,
has finite order. A similar argument shows that < gn > can’t contain g-1.) Every
element x ∈ G that is not a generator of
G is in S, since < x > is a proper subgroup. Thus G − S = {g, g-1}, so |G − S| =
2.
Answer: (b) As in (a), G − S is the set of elements of G which are
generators of G; if G is a finite cyclic group of order n ,
then it is isomorphic to
, and hence has
Ø(n) generators. So |G − S| = Ø(n).
Answer: (c) Since |G − S| > 0, we see that G is never the union S of its
proper subgroups.
[2] Let G be a group.
(a) Let F and H be subgroups of G, and as sume that F does not contain H and that
H does not contain F. Let f be
an element of F that is not in H and let h be an element of H that is not in F.
Show that fh is not in either F
nor H (i.e., show that fh is not in F ∪ H).
(b) Show that G is not the union of any two proper subgroups.
Answer: (a) If fh ∈ F, then fh = g for some g ∈ F, so h = f -1g ∈ F,
contradicting our assumption. Similarly, if fh ∈ H,
then f ∈ H, which is a contradiction. This means that fh is in neither F nor H.
Answer: (b) Say G were the union of two proper subgroups; call them F and
H. If F 
H, then G = F ∪ H = H, which
contradicts H being proper. Likewise , we can’t have H
F. Thus neither of F and H
contains the other, so there is an
f ∈ F − H and an h ∈ H − F, so fh is in neither F nor H, which means that G
can’t be the union of F and H.
[3] Let 
be the Fibonacci sequence (thus
and 
for every n ≥1). Prove that
for all n ≥ 5.
Answer: Clearly, 
, and
. And if 
and
for some k ≥ 6, then
. Now
for all n ≥ 5 fol lows by induction .
[4] Prove that 
is isomorphic to
, but not to
.
Answer: First, by the Chinese Remainder Theorem,
.
But no element of has order more than 60,
since 60 is the lcm of 12 and 30, whereas 
has order
120, so is not isomorphic to
.
[5] Let f : G → H be a homomorphism of groups.
(a) Define the kernel of f.
(b) Prove that the kernel of f is a subgroup of G.
(c) Prove that the kernel of f is a normal subgroup of G.
Answer: (a) ker f = {x ∈ G|f(x) = eH}
Answer: (b) Since eG ∈ ker f, we know ker f is not empty. If x, y ∈ker
f, then f(xy) = f(x)f(y) = eHeH = eH, so
xy ∈ ker f, so ker f is closed under the group ope ration . And if x ∈ ker f, then
, so
x-1 ∈ ker f, hence ker f is closed under taking inverses. Thus ker f is a
subgroup.
Answer: (c) Let x ∈ ker f and let g ∈ G. Then
, so
gxg-1∈ ker f, hence ker f is normal.
[6] Let n > 1 be a positive integer .
(a) Prove that the number of elements of order n in
is at least (n − 1)!.
[Hint: look at n-cycles.]
(b) Prove that has an element of order n that is not an n-cycle if and only
if n is not a power of a prime .
Answer: (a) There are n! ways to write down an n-cycle (since this is the number
of ways of ordering the numbers 1 to n).
But these can be grouped into sets of n orderings which define the same n-cycle,
so there are n!/n = (n −1)! n-cycles in .
Answer: (b) If n is not a power of a prime, then we can factor n so that
n = km, where 1 < k < m < n, gcd(k,m) = 1, but
n = km. Now k +m < 2m ≤ km = n, so we can find a disjoint k-cycle (call it σ) and
m-cycle (call it 
) in
. Then 
has
order n, since n is the lcm of k and m. Conversely, assume
has an element
of
order n but that is not an n-cycle. If
is a cycle, it must be an r-cycle with r < n, but then it has order r < n. Thus
is a product of disjoint cycles, and the lcm
of the lengths of the cycles is n. If n were a power of a prime, then since each
length divides n , the lengths are powers of the
same prime. Thus the lcm is the length which is the largest power, but all of
the lengths are less than n, so the order would
be less than n, contrary to assumption. Thus n can’t be a power of a prime.
