Download Solutions// Homework set 4

Solutions// Homework set 4
For this assignment, assume all problems are boldfaced. That is all require
complete sentence proofs with the exception of page 57, problem 39.
1. Show that if a subgroup H of Z8 (under addition modulo 8) contains
the element 3 = [3]8 , then H = Z8 .
Proof: Suppose H is a subgroup of Z8 and that 3 ∈ H. Then by
closure 3n ∈ H for all n ∈ Z (i.e., h3i ⊂ H). But this implies
3
6
1
4
7
2
5
=
=
=
=
=
=
=
3
3+3
3+3+3
3+3+3+3
−3 − 3 − 3
−3 − 3
−3
are all elements of H also. However, this implies that H = Z8 .
2. Find all subgroups of Z8 and draw the subgroup diagram (again under
addition modulo 8).
Solution: By class on Thursday, we know all subgroups are cyclic, and
if they are generated by a divisor of 8. This means all subgroups are
h0i ⊂ h4i ⊂ h2i ⊂ h1i = Z8 .
(Note, we can see that any subgroup that contains 1, 3, 5, 7 must be
all of Z8 . This follows since
Z8 = h1i = h−1i(= h7i),
and any subgroup containing 5 must contain −5 = 3 by 5.14(ii) and
then problem 1 implies that the subgroup is Z8 ).
3. Find all subgroups of U9 = {[1]9 , [2]9 , [4]9 , [5]9 , [7]9 , [8]9 } under multiplication (modulo 9).
Solution: We first calculate the cyclic subgroups.
h1i
h2i
h4i
h5i
h7i
h8i
=
=
=
=
=
=
{1}
{1, 2, 4, 8, 7, 5} = U9
{1, 4, 7}
{1, 5, 7, 8, 4, 2}
{1, 7, 4}
{1, 8}.
Using Thursday/s class, again, since U9 = h2i, we have that it is cyclic
and consequently all subgroups must be cyclic, so the above list is
complete (although redundant since h2i = h5i and h4i = h7i). If we
didn’t use Thursday’s class’s theorem, then we would have from the
above that no proper subgroup could contain 2 or 5, and we would be
left with worrying about {1, 4, 7, 8} (if a subgroup contains 4 or 7 it
contains the other since they are inverses). However, this subset cannot
be a subgroup since 4 ∗ 8 = 5 so it is not closed.
4. Use Theorem 5.14 to determine whether the following are subgroups of
Z (under addition).
(a) 14Z = {14n | n ∈ Z}.
This is a subgroup: We show this as follows. For (i), let a, b ∈
14Z. Then a = 14n1 and b = 14n2 for some integers n1 , n2 . Now
a + b = 14n1 + 14n2 = 14 ∗ (n1 + n2 ) ∈ 14Z.
This shows (i).
For (ii), we note that 0 = 14 · 0 ∈ 14Z since 0 ∈ Z.
Finally for (iii), let a ∈ 14Z. Then a = 14n for some n ∈ Z. But
now,
a−1 = −a = −14n = 14(−n) ∈ 14Z.
Hence 14Z satisfies the three conditions of 5.14 and is a subgroup.
(b) The set of all odd integers together with 0.
This set is not closed since 1 + 1 = 2 but 2 is not an odd integer.
5. Determine (with proof) whether the following are subgroups of Q] under multiplication.
(a) { pq | p and q are primes}.
This is not a subgroup since
2 5
10
· = .
3 7
21
However, 10 and 21 are not primes. Moreover since 10
is in lowest
21
terms, there is no way to write it as a quotient of primes. Hence
this set is not closed and (i) is violated. Therefore it is not a
subgroup.
(b) H = {2m 3n | m, n ∈ Z}.
This is a subgroup: To see this, we begin by checking (i). Suppose a, b ∈ H. Then a = 2m1 3n1 for some integers m1 , n1 , and
b = 2m2 3n2 for some integers m2 , n2 . By arithmetic,
ab = 2m1 +m2 3n1 +n2 ,
which is an element of H since m1 + m2 and n1 + n2 are both
integers. Hence H is closed.
For (ii), note that 1 is the identity of H. Since 1 = 20 30 , then
1 ∈ H.
For (iii), suppose a ∈ H. Then a = 2m 3n for some integers m, n.
Now, a−1 = 2−1 3−1 , so a−1 ∈ H. Hence H satisfies (i), (ii), (iii)
of 5.14 showing that H is a subgroup.
6. Let G be a group. We define the center of G, to be the set
Z(G) = {g | gx = xg for all x ∈ G}.
Prove that Z(G) is a subgroup of G.
Proof: We will show the three conditions of theorem 5.14. Note that
for convenience, we write g1 g2 to denote g1 ∗ g2 , where ∗ is the binary
operation of G. For (i), suppose g1 , g2 ∈ Z(G). Then for all x ∈ G,
we have xg1 = g1 x and xg2 = g2 x. Let a = g1 g2 , and let x ∈ G be
arbitrary. Then
ax =
=
=
=
=
=
(g1 g2 )x
g1 (g2 x)
g1 (xg2 )
(g1 x)g2
(xg1 )g2
xa
by
by substitution
by associativity
since xg2 = g2 x
by associativity
since xg1 = g1 x
substitution.
Since x ∈ G was arbitrary, a ∈ Z(G) showing that Z(G) is closed under
the operation of G.
For (ii), we note that if e is the identity of G and x ∈ G is given, then
by definition ex = x = xe. Consequently e ∈ Z(G).
For (iii), Suppose a ∈ Z(G) and x ∈ G is arbitrary. Then ax =
xa. Multiplying by a−1 on both the right and left hand sides of this
equation, we obtain the following sequence of equations:
(a−1 (ax))a−1
((a−1 a)x)a−1
(ex)a−1
xa−1
=
=
=
=
(a−1 (xa))a−1
a−1 (x(aa−1 ))
by associativity
−1
a (xe)
by definition of a−1
a−1 xby definition of e.
Since x ∈ G was arbitrary, it follows that a−1 ∈ Z(G) establishing (iii).
Therefore Z(G) satisfies conditions (i), (ii), and (iii) of 5.14, implying
that Z(G) is a subgroup.
7. Suppose G1 and G2 are both groups, and suppose that φ : G1 → G2
is an isomorphism of groups (and therefore of binary systems). Prove
that H1 is a subgroup of G1 if and only if H2 = {φ(h) | h ∈ H1 } is a
subgroup of G2 .
Proof: Let φ, G1 , G2 be as above, and assume that H1 ⊂ G1 and
H2 = φ(H1 ). We first suppose that H1 is a subgroup of G1 . We need
to check the three conditions of 5.14 for H2 . Let x, y ∈ H2 be given.
To see that x ∗0 y ∈ H2 , we need to find an element of h ∈ H1 such
that φ(h) = x ∗0 y. Now, there exist a, b ∈ H1 such that x = φ(a) and
y = φ(b). We now have
x ∗0 y = φ(a) ∗0 φ(b)
= φ(a ∗ b)
since φ is an isomorphism.
Now h = a ∗ b ∈ H1 since H1 is a subgroup (5.14 (i)). Thus x ∗0 y ∈ H2
since x ∗0 y = φ(h).
For (ii), note that if e1 is the identity element of G1 , by previous theorem, it follows that φ(e1 ) is the identity element of G2 . Now, as H1 is
a subgroup, by 5.14(ii), e1 ∈ H1 . But then e2 = φ(e1 ) ∈ H2 as desired.
For (iii), suppose x ∈ H2 . Then x = φ(a) for some a ∈ H1 . By 5.14(iii),
a−1 ∈ H1 . We now look at
φ(a) ∗0 φ(a−1 ) = φ(a ∗ a−1 )
since φ is an isomorphism
= φ(e1 )
by definition of a−1
= e2
with the last following by the discussion for (ii). But now, this implies
that x−1 = φ(a)−1 = φ(a−1 ) ∈ H2 . Since x ∈ H2 was arbitrary, it
follows that H2 satisfies (iii). Thus by 5.14, H2 is a subgroup.
For the converse, we note that if φ is an isomorphism, then φ−1 : G2 →
G1 exists. Moreover, for x, y ∈ G2 , we have that x = φ(a) for some
a ∈ G1 and y = φ(b) for some b ∈ G1 . Thus
φ−1 (x ∗0 y) =
=
=
=
φ−1 (φ(a) ∗0 φ(b))
by substitution
−1
φ (φ(a ∗ b))
since φ is an isomorphism
a∗b
by definition of φ−1
φ−1 (x) ∗0 φ−1 (y)
by definition of φ−1 .
Thus, φ−1 is also an isomorphism. But now, we can apply the first half
of the problem, interchanging the roles of H1 and H2 .
8. Let G be a group and suppose H and K are both subgroups of G.
Prove that H ∩ K is also a subgroup of G.
Proof: We again work by showing the three conditions of 5.14.
For (i), suppose a, b ∈ H ∩ K. By definition of the intersection a, b ∈ H
and a, b ∈ K. Since H and K are subgroups, it follows by 5.14(i)
applied to H and K that ab ∈ H and ab ∈ K. Hence, ab ∈ H ∩ K, so
that H ∩ K satisfies 5.14(i).
For (ii), note that by 5.14(ii), e ∈ H and e ∈ K. Hence e ∈ H ∩ K
establishing (ii) for H ∩ K.
For (iii), we let a ∈ H ∩ K. As before, a ∈ H and a ∈ K. But applying
5.14(iii) to H and K, we have a−1 ∈ H and a−1 ∈ K. Hence it follows
that a−1 ∈ H ∩ K.
Consequently H ∩ K satisfies all three conditions of 5.14, and it follows
that H ∩ K is a subgroup.
9. page 57, problem 39.
(a) True, (b) False, (c) True, (d) False G is the only improper subgroup
of G. The identity subgroup is the trivial subgroup, (e) False (the
identity is never a generator if the group has more than one element)
(f) False (In Z3 for example, both 1 and 2 are generators), (g) False
(almost never true since if there is more than one element, the additive
identity 0 has no inverse - be careful, though {0} is a group under
multiplication since 0 is then the identity), (h) False, it must also have
the three properties of 5.14. Note the counterexamples in problem 4
and 5, (i) True it is generated by [1]4 , (j) False, again we need the
properties of 5.14 and problems 4 and 5 provide counterexamples.
10. Find all generators for the cyclic group U7 = {[1]7 , [2]7 , . . . , [6]7 } under
multiplication.
We calculate all cyclic subgroups and obtain:
h1i
h2i
h3i
h4i
h5i
h6i
=
=
=
=
=
=
{1}
{1, 2, 4}
{1, 3, 2, 6, 4, 5}
{1, 4, 2}
{1, 5, 4, 6, 2, 3}
{1, 6}
So the generators are 3 and 5.
11. Give an at most three sentence proof synopsis of the proof of theorem
6.6. If G = hai is a cyclic group and H 6= hei is a non-trivial subgroup,
we take the smallest positive power k of a such that ak ∈ H. Using the
division algorithm, we show that if another element am is in H, then
the remainder of the m upon division by k is 0, so that m is a multiple
of k. Consequently, by the exponent rules am is a power of ak and ak
generates H.