Assignment 3b
28, 29, 32, 49, 50, 67, 70, 72, 73
28. Suppose that H is a proper subgroup of Z under addition and that H contains 18, 30, and 40.
Determine H.
Since H is a subgroup it is closed under addition and subtraction (additive inverses) so we see
that 30 − 18 = 12 ∈ H, 40 − 30 = 10 ∈ H, and then 12 − 10 = 2 ∈ H. Then all multiples of 2 (all
even numbers) must be contained in H. If any odd number k ∈ H, then k + 1 is a multiple of 2
so k + 1 ∈ H and hence (k + 1) − k = 1 ∈ H. This would imply that H = Z, which contradicts
that H is a proper subgroup. Therefore H contains no odd integer, so H is the set of all even
integers: H = {2k | k ∈ Z}.
29. Suppose that H is a proper subgroup of Z under addition and that H contains 12, 30, and 54.
What are the possibilities for H?
As in problem 28, we see that H must contain 30 − 18 = 12 and also 18 − 12 = 6. Then H must
contain all multiples of 6. If H contains any number that is not a multiple of 6, by subtracting
we can see that H contains a number between 1 and 5.
If
If
If
If
If
1 ∈ H,
2 ∈ H,
3 ∈ H,
4 ∈ H,
5 ∈ H,
then
then
then
then
then
H
H
H
H
H
= Z and is not a proper subgroup.
could be the set of evens, since the multiples of 6 are all even.
could be the set of multiples of 3, since multiples of 6 are also multiples of 3.
contains 6 − 4 = 2, so H could be the evens.
contains 6 − 5 = 1 and is not proper.
Therefore H could be {2k | k ∈ Z}, {3k | k ∈ Z}, or {6k | k ∈ Z}.
32. If H and K are subgroups of G, show that H ∩ K is a subgroup of G.
Proof. Let H, K be subgroups of G. We will use the 2-Step Subgroup Test to show that H ∩ K
is a subgroup of G.
Since e ∈ H and e ∈ K, e ∈ H ∩ K so H ∩ K 6= ∅.
Let a, b ∈ H ∩ K. Thus a, b ∈ H and a, b ∈ K. Since a, b are both in H and H is closed we see
that ab ∈ H. Similarly ab ∈ K. Thus ab ∈ H ∩ K.
Let a ∈ H ∩ K. Thus a ∈ H and a ∈ K. Since a ∈ H and H is closed under inverses, a−1 ∈ H.
Similarly, since a ∈ K, a−1 ∈ K. Thus a−1 ∈ H ∩ K.
Therefore by the 2-Step Subgroup Test, H ∩ K is a subgroup of G.
49. Suppose a group contains elements a and b such that |a| = 4, |b| = 2, and a3 b = ba. Find |ab|.
Starting with a3 b = ba and multiplying on the left by a and then on the right by b we get
a3 b = ba
a4 b = aba
eb = aba
b = aba
b2 = abab
e = (ab)2
So |ab| ≤ 2. The only way |ab| = 1 is if ab = e and thus a = b−1 . By problem 4 above, inverses
must have the same order. But |a| = 4 6= 2 = |B|, which is a contradiction. Thus |ab| = 2.
50. Suppose a and b are group elements such that |a| = 2, b 6= e, and aba = b2 . Determine |b|.
Since b 6= e, |b| =
6 1.
If |b| = 2, then multiplying the equation aba = b2 on the left by a and on the right by a gives:
aba = b2
aba = e
2 2
a ba = aea
ebe = a2
b=e
which is a contradiction. So |b| =
6 2.
Now consider b4 = (b2 )2 = (aba)2 = abaaba = abeba = a(b2 )a = a(aba)a = a2 ba2 = ebe = b.
Then multiplying by b−1 gives b3 = e. Thus |b| = 3.
a b
a b
67. Let G =
a, b, c, d ∈ Z under addition. Let H =
∈G a+b+c+d=0 .
c d
c d
Prove that H is a subgroup of G. What if 0 is replaced by 1?
0 0
Proof. We will use the 1-Step Subgroup Test. Note that H is nonempty since
∈ H.
0 0
a1 a2
b1 b2
Let A =
∈ H and B =
∈ H. Thus a1 +a2 +a3 +a4 = 0 and b1 +b2 +b3 +b4 = 0.
a3 a4
b
3 b4
a1 − b 1 a2 − b 2
Then A − B =
. Adding all of the entries gives
a3 − b 3 a4 − b 4
(a1 − b1 ) + (a2 − b2 ) + (a3 − b3 ) + (a4 − b4 ) = (a1 + a2 + a3 + a4 ) − (b1 + b2 + b3 + b4 ) = 0 − 0 = 0.
Thus A − B ∈ H and H ≤ G by the 1-Step Subgroup Test.
70. Let G be a group of functions from R to R∗ , where the operation of G is multiplication of
functions. Let H = {f ∈ G | f (2) = 1}. Prove that H is a subgroup of G. Can 2 be replaced by
any real number?
Proof. We will use the 2-Step Subgroup Test. Note that the constant function f (x) = 1 is in H
so H 6= ∅.
Let f, g ∈ H. Then f (2) = 1 and g(2) = 1 so f g(2) = f (2)g(2) = 1 · 1 = 1, and thus f g ∈ H.
1
1
Let f ∈ H. Then f (2) = 1. The inverse of f is (f (x))−1 = f (x)
since [f (f )−1 ](x) = f (x) f (x)
= 1,
−1
and the constant function 1 is the multiplicative identity (note that (f ) is defined since f (x) 6= 0
as 0 is not in the codomain R∗ ). Since (f (2))−1 = 1/f (2) = 1/1 = 1, we see that (f )−1 ∈ H.
Therefore by the 2-Step Subgroup Test H is a subgroup of G.
Yes, 2 can be replaced by any real number. As seen in the proof, there is nothing special about
2, just the fact that the same number is always used for evaluating.
72. Let H = {a + bi | a, b ∈ R, ab ≥ 0}. Prove or disprove that H is a subgroup of C under addition.
H is not a subgroup since it is not closed:
2 + i ∈ H since 2 · 1 = 2 > 0 and −1 − 2i ∈ H since (−1)(−2) = 2 > 0.
But (2 + i) + (−1 − 2i) = 1 − i 6∈ H since 1(−1) = −1 6≥ 0.
73. Let H = {a + bi | a, b ∈ R, a2 + b2 = 1}. Prove or disprove that H is a subgroup of C∗ under
multiplication. Describe the elements of H geometrically.
H is a subgroup of C∗ .
Proof 1. We will use the 2-Step Subgroup Test. H is nonempty since 1 = 1 + 0i ∈ H.
Let z1 , z2 ∈ H. Then z1 = a1 + b1 i, where a21 + b21 = 1 and z2 = a2 + b2 i where a22 + b22 = 1.
Then z1 z2 = (a1 + b1 i)(a2 + b2 i) = a1 a2 − b1 b2 + (a1 b2 + a2 b1 )i. Now consider
(a1 a2 − b1 b2 )2 + (a1 b2 + a2 b1 )2 = (a21 a22 − 2a1 a2 b1 b2 + b21 b22 ) + (a21 b22 + 2a1 a2 b1 b2 + a22 b21 )
= a21 a22 + a21 b22 + b21 b22 + a22 b21
= a21 (a22 + b22 ) + b21 (b22 + a22 )
= a21 (1) + b21 (1)
=1
So z1 z2 ∈ H.
Now for z = a + bi ∈ H with a2 + b2 = 1, z −1 =
a2 + (−b)2 = a2 + b2 = 1, we see that z −1 ∈ H.
1
z
=
1
a+bi
· a−bi
=
a−bi
a−bi
a2 +b2
=
a−bi
1
= a − bi and since
Therefore by the 2-Step Subgroup Test, H is a subgroup of C∗ .
Proof 2. We will use the 1-Step Subgroup Test. H is nonempty since 1 = 1 + 0i ∈ H.
p
Let z1 , z2 ∈ H. Then z1 = a1 + b1 i = r1 eiθ1 , where r1 = a21 + b21 and arg(z1 ) (recall tan θ =
b1 /a1 ). Since z1 ∈ H, we have a21 + b21 = 1, so r1 = 1 and z1 = eiθ1 . Similarly z2 = eiθ2 . The
inverse of z2 is z2−1 = z12 = e−iθ2 .
Then z1 z2−1 = eiθ1 e−iθ2 = ei(θ1 −θ2 ) , and |z1 z2−1 | = 1. So z1 z2−1 ∈ H.
Therefore by the 1-Step Subgroup Test, H is a subgroup of C∗ .
The elements of H are points in the unit circle in the complex plane.