Franz-Viktor Kuhlmann
210 McLean Hall, Tel. 6111
Abstract Algebra — Math 363.3 (02) T2, 2004-2005
Problem set 6 – Some sample solutions
2) U (15) = {1, 2, 4, 7, 8, 11, 13, 14}. The cyclic subgroups generated by these elements
are:
h2i = h8i = {1, 2, 4, 8},
h4i = {1, 4},
h7i = {1, 4, 7, 13},
h11i = {1, 11} (note that 11 = −4),
h14i = {1, 14} (note that 14 = −1),
h1i = {1}.
Observe that U (15) cannot be cyclic since it has three subgroups of order 2 and two of
order 4.
3) Suppose a is an element of order 3. Then the group it generates has three elements,
one of which is the neutral element 1. The others are a and a2 . Since (a2 )2 = a4 = a 6= 1
and (a2 )3 = (a3 )2 = 1, we see that the order of a2 is 3. So every subgroup of order
3 contains exactly two elements of order 3. Moreover, if a or a2 lies in any subgroup
of order 3, then this subgroup must be equal to hai = ha2 i. This shows that the eight
elements of order 3 are distributed over four cyclic subgroups of order 3, two of them
in each subgroup. By problem 12 below, every subgroup of order 3 is cyclic and must
therefore contain one (and actually, two) of the elements of order 3. This shows that G
has exactly four subgroups of order 3, and they are all cyclic.
4) If a subgroup of order 4 contains an element of order 4, then it is cyclic. So we have
to look for elements of order less than 4 in U (40). Checking numbers of the form 1 + 40k
whether they are squares gives 81 = 92 and 121 = 112 . Thus, 9 and 11 are elements of
2
order 2 in U (40). We find 9 · 11 = 19 which is also of order 2 since 19 = (9 · 11)2 =
2
2
2
2
9 · 11 = 1. Since 9 · 19 = 9 · 11 = 11 and 11 · 19 = 9 · 11 = 9 and U (40) is commutative,
we see that {1, 9, 11, 19} is closed under multiplication and thus a subgroup of U (40) of
order 4. It is not cyclic since none of its elements has order 4.
5) By assumption, G is a group of functions from R to R \ {0} under multiplication of
functions, that is, if f, g ∈ G then the product f · g is defined by (g · h)(x) = g(x) · h(x).
The neutral element is the constant function f (x) = 1 because for every function g ∈ G,
(f · g)(x) = 1 · g(x) = g(x) and (g · f )(x) = g(x) · 1 = g(x), so f · g = g and g · f = g.
Since f (1) = 1, f is an element of H and therefore, H 6= ∅. Now we use the Two Step
Subgroup Criterion. Take any two functions g, h ∈ H. Then g(1) = 1 and h(1) = 1, hence
(g · h)(1) = g(1) · h(1) = 1 · 1 = 1, which shows that g · h ∈ H. The inverse of a function
1
; this is possible because by assumption, all
g ∈ G is the function h defined by h(x) = g(x)
1
functions in G take values in R\{0}. Indeed, (g·h)(x) = g(x)·h(x) = g(x)· g(x)
= 1 = f (x),
1
= 11 = 1,
so g · h = f , and the same holds for h · g. Now if g ∈ H, then we have h(1) = g(1)
which shows that the inverse h of g is also in H.
1
6) If |a| = n is finite, then by Corollary 2, page 77, hai = hak i if and only if gcd(k, n) = 1.
Hence if |a| = 6, then the only generators of hai are a and a5 = a−1 . If |a| = 8,
then the generators of hai are a, a3 , a5 , a7 . If |a| = 8, then the generators of hai are
a, a3 , a7 , a9 , a11 , a13 , a17 , a19 .
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8) The subgroup of order 8 in G is generated by a 8 = a3 (Theorem 4.3). Since |a3 | = 8,
the element (a3 )k in ha3 i is a generator of ha3 i if and only if gcd(k, 8) = 1 (Corollary 2,
page 77). This holds for k = 1, 3, 5, 7. So the generators of the subgroup of order 8 are
(a3 )1 = a3 , (a3 )3 = a9 , (a3 )5 = a15 , and (a3 )7 = a21 .
9) Since a has infinite order, the same holds for a3 since if it would have order n < ∞,
then 1 = (a3 )n = a3n and a would have finite order ≤ 3n. By Theorem 4.1, it follows that
all distinct powers (a3 )k of a3 are distinct group elements. If also (a3 )m is a generator
of ha3 i, then ((a3 )m )` = a3 for some ` ∈ Z since a3 ∈ ha3 i = h(a3 )m i. So it follows that
3 = 3m`, that is, m` = 1. Since m, ` ∈ Z, this is only possible if m = 1 or m = −1. So
the only generators for ha3 i are a3 and a−3 .
10) Suppose the element ak in the cyclic group hai is of finite order, say, n. Then
akn = (ak )n = 1. If kn is positive, then we have a contradiction because a has infinite
order and therefore, akn 6= 1. If k is negative, then −k is positive, and we compute:
a−kn = (akn )−1 = 1−1 = 1. Since now −kn is positive, we get again a contradiction. So
kn must be zero. Since n ≥ 1, this means that k = 0, showing that the only element of
finite order in an infinite cyclic group is the neutral element a0 = 1.
11) We note that an−k = an a−k = 1 · a−k = a−k is the inverse of ak . So we actually have
to show that the inverse of an element ak in hai has the same order as ak .
One way to show this is to observe that if the order of an element b is m, then
−1 m
(b ) = b−m = (bm )−1 = 1−1 = 1, so |b−1 | ≤ m = |b|. But since b is the inverse of b−1 ,
we also get |b| ≤ |b−1 |. So we have |b| = |b−1 |.
Another way to show this is by using that gcd(n, −k) = gcd(n, k), hence by Theorem
4.2,
n
n
|a−k | =
=
= |a−k | .
gcd(n, −k)
gcd(n, k)
12) Let G be a group of order 3. If it contains an element a of order 3, then hai is a
subgroup of G of order 3, hence G = hai and G is cyclic. Now suppose G does not contain
an element of order 3, and consider the two elements a, b ∈ G which are different from the
neutral element. Then their order is not 1. Their order cannot be bigger than 3 because
then they would generate subgroups that have more elements than G, which is absurd.
So their order must be 2. Since G has only the three elements 1, a, b, we have ab = 1 or
ab = a or ab = b. But ab = a would imply b = 1 and ab = b would imply a = 1, so we
can only have that ab = 1. That is, b is the inverse of a, and so it must be an element
of the subgroup hai = {1, a}. Since b 6= 1 and b 6= a, this is a contradiction. This shows
that both a and b must have order 3.
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