Math 307 Abstract Algebra Homework 3
Sample Solution
1. Let H = {a + bi : a, b ∈ R, ab ≥ 0}. Prove or disprove that H is a subgroup of C under
addition.
Solution. Note that 1, −i ∈ H, but 1 + (−i) ∈
/ H. So, H is not a subgroup.
2. Let H be a non-trivial subgroup of Z. Then H has some nonzero elements a so that a, −a ∈ H,
and one of them is positive. As a result, the set of positive numbers in H is non-empty. By
the well ordering property of H, there is a smallest positive integer k in H. We claim that
every element x ∈ H is an integral multiple of k. It then follows that H = kZ = hki.
Suppose our claim is not true. Then there is h = kq + r ∈ H with 0 < r < k. Note that
k, h ∈ H so that r = h − qk = h − k − k − · · · − k ∈ H, which contradicts the assumption that
k is the smallest positive integer in H. Our claim follows.
3. Let H be a non-trivial subgroup of Zn . Suppose H has elements h̄1 , . . . , h̄m = 0̄ with 0 <
h1 < h2 < · · · < hm = n so that h1 is the smallest positive integer satisfy h̄1 ∈ H. We claim
that hs = `s h1 = h1 + · · · + h1 (hs times) for some positive integer `s for s = 2, . . . , m. It will
then follow that H = hh1 i.
Suppose our claim is not true. Then there is hs = `s h1 + r with 0 < r < h1 . Note that
h̄s , h̄1 ∈ H so that r̄ = h̄s − `s h̄1 = h̄s − h̄1 − h̄1 − · · · − h̄1 ∈ H, which contradicts the
assumption that h1 is the smallest positive integer satisfying h̄1 ∈ H.
4. Determine the subgroup lattice of Z8 .
Solution. By checking gcd(8, k) for k = 1, . . . , 7, we see that there are four subgroups in Z8 :
h1i = h3i = h5i = h7i,
h2i = h6i,
The lattice diagram (in horizontal form) is:
h4i,
h0i.
h1i − h2i − h4i − h0i.
5. Let a and b be elements of a group such that |a| = 4, |b| = 2, and a3 b = ba. Find |ab|.
Solution. We prove that |ab| = 2. Note that (ab)(ab) = a(ba)b = a(a3 b)b = a4 b2 = e. So,
|ab| = 1 or 2. If |ab| = 1, then a is the inverse of b so that 4 = |a| = |b| = 2, which is absurd.
So, |ab| = 2.
6. Suppose G is a group with n elements, and H is a subgroup of G with m elements.
(a) Suppose H 6= G and g1 ∈ G − H. Let g1 H = {g1 h : h ∈ H}.
Show that H ∩ g1 H = ∅ so that |H ∪ g1 H| = 2m. (Here you need to argue |g1 H| = m.)
(b) Suppose H ∪ g1 H 6= G and g2 6∈ (H ∪ g1 H).
Show that (H ∪ g1 H) ∩ g2 H = ∅ so that |H ∪ g1 H ∪ g2 H| = 3m.
(c) Show that G is a disjoint union of H ∪ g1 H ∪ g2 H · · · gk H for some g1 , . . . , gk ∈ G
so that n/m is a positive integer.
Solution. Let H = {h1 , . . . , hm } with hm = e.
(a) Let g1 ∈ G − H, and g1 H = {g1 h1 , . . . , g1 hm }. We claim that H ∩ g1 H = ∅. If it is not
true, then there are i, j such that g1 hi = hj so that g = hj h−1
i ∈ H, which is a contradiction.
1
Note that if g1 hi = g1 hj , then hi = hj by the cancellation law. Thus, g1 H has m elements,
and H ∪ g1 H has 2m elements.
(b) If there is g2 ∈ G − (H ∪ g1 H), let g2 H = {g2 h1 , . . . , g2 hm }. We claim that g2 H ∩ (H ∪
g1 H) = ∅. If not, g2 hi = hj or g2 hi = g1 hj for some i, j. Thus, g2 = hj h−1
or g2 = g1 hj h−1
i
i
so that g2 ∈ H ∪ g1 H. Now, g2 H has m elements so that H ∪ g1 H ∪ g2 H has 3m elements.
(c) We can repeat the arguments in (a) and (b) as follows. If H, g1 H, . . . , gr H are constructed
and their union has (r + 1)m elements, and if gr+1 ∈ G − (H ∪ g1 H ∪ · · · ∪ gr H), then
gr+1 H = {gr+1 h1 , . . . , gr+1 hm } are different from those in H ∪ g1 H ∪ · · · ∪ gr H. If not,
then gr+1 hi = hj or gr+1 hi = g` hj for some ` ∈ {1, . . . , r}. But then gr+1 = hj h−1
or
i
gr+1 = g` hj h−1
∈
g
H,
which
contradicts
the
fact
that
g
∈
G
−
(H
∪
g
H
∪
·
·
·
∪
g
H).
r+1
1
r
`
i
Hence, we can repeat this process, and construct H, g1 H, g2 H, . . . , gk H until all the elements
in G are exhausted. It follows that n = (k + 1)m.
7. (a) Suppose a group G has order p, which is a prime. Then p ≥ 2 and there is a ∈ G not
equal to e. Then H = hai ≤ G, and H has order larger than one. By Problem 6, |H| is a
factor of p and not equal to 1. Thus, |H| = p and G = H = hai is cyclic.
(b) Let a, b ∈ G. Suppose |a| = n and |b| = m such that gcd(m, n) = 1. Let H = hai ∩ hbi
has order k. Then H ≤ hai implies that k is a factor of n, and H ≤ hbi implies that k is a
factor of m. Hence, k is a common divisor of n and m. Thus, k = 1 and H = {e}.
8. Suppose |G| = 24 and G is cyclic. If a8 6= e and a12 6= e, show that G = hai.
Solution. Note that the order of a ∈ G must be a factor of 24 so that |a| ∈ {1, 2, 3, 4, 6, 8, 12, 24}.
If |a| = 1, 2, 4, 8,, then a8 = e; if |a| = 3, 6, 12,, then a12 = e. Thus, |a| = 24 and G = hai.
Extra credits
9. Suppose G is a set equipped with an associative binary operation ∗. Furthermore, assume
that G has an left identity e, i.e., eg = g for all g ∈ G, and that every g ∈ G has an left
inverse g 0 , i.e., g 0 ∗ g = e. Show that G is a group.
Solution. Let g ∈ G. We first show that the left inverse g 0 of g is also the right inverse, i.e.,
g ∗ g 0 = e. Note that e = (g 0 )0 ∗ g 0 = (g 0 )0 ∗ (e ∗ g 0 ) = (g 0 )0 ∗ (g 0 ∗ g ∗ g 0 ) = e ∗ (g ∗ g 0 ) = g ∗ g 0 .
To show that the left identity is also the right identity, observe that g ∗ g 0 = g 0 ∗ g = e for any
g ∈ G by the proof in the preceding paragraph. So, we have g ∗ e = g ∗ (g 0 ∗ g) = (g ∗ g 0 ) ∗ g = g.
10. Let A be a set, and P(A) be its power set. Show that there is a group G with |G| = |P(A)|,
Proof. Case 1. If |A| = n is finite, then ZN with N = 2n is a group with 2n elements, where
|ZN | = |P(A)|.
Case 2. A is infinite. Let SA be the group of bijections (permutations) on A under function
composition. By the Axiom of Choice, we have |A × A| = |A| so that |P(A × A)| = |P(A)|.
Clearly, every bijection on A corresponds to a subset of A × A. So, there is an injection from
SA to P(A × A), i.e., |SA | ≤ |P(A × A)| = |P(A)|.
Now, for every subset S of A, there is a bijection f : A → A such that f (x) = x for all x ∈ S
and f (x) 6= x for all x ∈
/ S. Thus, there is an injection from P(A) to SA , i.e., |P(A)| ≤ |SA |.
By the Schroder-Berstein Theorem, |P(A)| = |SA |.
2