# Math 307 Abstract Algebra Homework 3 Sample Solution

```Math 307 Abstract Algebra Homework 3
Sample Solution
1. Let G be a group and a ∈ G.
(1) Show that C(a) = {g ∈ G : ag = ga} is a subgroup of G.
(2) Show that Z(G) = ∩a∈G C(a).
Solution. (1) Let a ∈ G. Clearly, e ∈ C(a) because ae = ea. If x, y ∈ C(a), then (xy)a =
x(ya) = x(ay) = (xa)y = a(xy) so that xy ∈ C(a). Also, xa = ax implies that ax−1 =
x−1 (xa)x−1 = x−1 (ax)x−1 = x−1 a. So, C(a) is a subgroup.
(2) Note that x ∈ Z(G) if and only if xa = ax for all a ∈ G, i.e., x ∈ C(a) for each a ∈ g.
2. Let H = {a + bi : a, b ∈ R, ab ≥ 0}. Prove or disprove that H is a subgroup of C under
Solution. Note that 1, −i ∈ H, but 1 + (−i) ∈
/ H. So, H is not a subgroup.
3. Determine the subgroup lattice of Z8 .
Solution. Note that h1i = h3i = h5i = h7i h2i = h6i, and
h1i − h2i − h4i − h0i.
4. Suppose that a group contains elements a and b 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.
5. Let a and b belong to a group. If |a| and |b| are relatively prime, show that hai ∩ hbi = {e}.
Solution. Suppose H = hai = {a, a2 , . . . , am } and K = hbi = {b, b2 , . . . , bn }, where am = bn =
e, such that gcd(m, n) = 1. Clearly, e ∈ H ∩ K. Suppose c ∈ H ∩ K and |c| = k. Then k is
factor of m and also a factor of n. Thus, k = 1 and c = e.
6. let G be a group and let a be an element of G.
(1) if a12 = e, what can we say about the order of a?
(2) if am = e, what can we say about the order of a?
(3) suppose that |G| = 24 and that G is cyclic. If a8 6= e and a12 6= e, show that G = hai.
Solution. (1) Then |a| = 1, 2, 3, 4, 6 or 12.
(2) Then |a| is a factor of |m|.
(3) By (2), |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.
7. Show that a group of prime order must be cyclic.
Solution. Suppose |G| = p is a prime. Then p > 1 and there is g ∈ G not equal to e. Suppose
H = hgi = {g, g 2 , . . . , g m } ≤ G has order m.
If H = G, then we are done. We show that it is impossible to have other elements in G−H. If
it did, and a1 ∈ G−H, then we claim that for a1 H = {a1 g, . . . , a1 g m } the set H ∪a1 H has 2m
elements. To see this, note that a1 g i 6= a1 g j for 1 ≤ i < j ≤ m, else g i = aj by cancellation.
Also, a1 g i 6= g j for any 1 ≤ i, j ≤ m, else a1 = g j−1 ∈ H, which is a contradiction. Now,
|G| = p 6= 2m = |H ∪ a1 H| So, there is a2 ∈ G − (H ∪ a1 H).
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In general, suppose Sk = H ∪ a1 H ∪ · · · ak H has (k + 1)m elements so that G 6= Sk and
ak+1 ∈ G − Sk . We show that Sk+1 = Sk ∪ ak+1 H has (k + 2)m elements by induction
on k. (When k = 1, the result holds by the discussion in the previous paragraph.) First
ak+1 g i 6= ak+1 g j for 1 ≤ i < j ≤ m, else ai = aj by cancellation. Next, ak+1 g i 6= ar g j for
any 1 ≤ i, j ≤ m and r ∈ {0, 1, . . . , k} with a0 = e, else ak+1 = ar g j−i ∈ Sk . So, our claim is
proved.
Now, we can repeat the above arguments to show that |G| = p > mn for any n ∈ N with
n > 1, which is a contradiction.
Extra credits
1. It is known that “For a group G, H, K ≤ G are such that H ∪ K ≤ G if and only if H ≤ K
or K ≤ H.” Find a suitable extension of the following statement to an arbitrary collection
of subgroups of G, or explain why it is not possible.
Solution. If one of the subgroup contains all other, then the union is clearly a subgroup.
But the converse is not true if there are more than 2 subgroups involved. For example, suppose
G = U (8) = {1, 3, 5, 7} ⊆ Z8 under multiplication. Let H1 = h3i = {3, 1}, H2 = h3i = {5, 1},
H3 = h3i = {7, 1}. Then H1 ∪ H2 ∪ H3 = G is a subgroup.
2. 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. To
see this, let gˆ be the left inverse of g 0 . Then (ˆ
g ) = (ˆ
g )(g 0 g) = (ˆ
g g 0 )g = eg = g. So, gˆ = g
0
0
satisfies e = gˆg = gg .
Now, because gg 0 = g 0 g = e, we have ge = g(g 0 g) = (gg 0 )g = g.
3. Consider the set R∗ of nonzero real number with the binary operation ab = |a|b. Show that
the operation is associative, R∗ has a left identity, and every a ∈ R∗ has a right inverse. But
R∗ is not a group under this operation.
Solution. Let a, b, c ∈ R∗ . Then clearly the operation is binary, and (ab)c = (|a|b)c = ||a|b|c =
|ab|c = |a||b|c = |a|(|b|c) = a(bc) so that it is associative.
Let e = 1. Then eb = b for all b ∈ R∗ . Also, for any a ∈ R∗ we can let a0 = 1/|a| so that
aa0 = 1.
But, f = −1 also satisfies f b = b for all b ∈ R∗ , and f (−1) = e(−1) does not imply f = e,
i.e., cancellation law fails. So, R∗ is not a group under the operation ab = |a|b.
4. Suppose x is an element of a cyclic group of order 15 and exactly two of x3 , x5 , and x9 are
equal. Determine |x13 |.
Solution. Let x ∈ G = hai = {a, . . . , a15 }. Clearly, |x| > 1, else e = x3 = x5 = x9 . Note also
that |x| is a factor of |G| = 15. Thus, |x| ∈ {3, 5, 15}. Consider 3 cases.
1. x3 = x5 6= x9 . Then x2 = x5−3 = e. So, |x| = 2, a contradiction.
2. x3 6= x5 = x9 . Then x4 = x9−5 = e so that |x| ∈ {2, 4}, a contradiction.
3. x3 = x9 6= x5 . Then x6 = x9−3 = e so that |x| ∈ {2, 3, 6}. Thus, |x| = 3 and |x13 | = |x| = 3.
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