MATH 103A: MODERN ALGEBRA I (WINTER 2014)
HW3 SOLUTIONS
Problem 1. Let G be a group with the property that ∀ x, y, z ∈ G, xy = zx implies y = z. Let
a, b ∈ G. To prove that G is Abelian, we will show that ab = ba.
Since G is a group and a ∈ G, the inverse a−1 exists and is an element of G. Then,
a−1 ( ab) = ( a−1 a)b = eb = b,
and
(ba) a−1 = b( aa−1 ) = be = b.
Thus, a−1 ( ab) = b = (ba) a−1 . By the property of G, this implies ab = ba.
Problem 2. U (12) = {1, 5, 7, 11}. The Cayley table for U (12) under multiplication mod 12 is:
.
1
5
7
11
1
1
5
7
11
5
5
1
11
7
7
7
11
1
5
11
11
7
5
1
Problem 3. Let G be the set of all rational numbers of the form 2m 3n where m and n are integers.
Then G is a group under multiplication, as followed:
• Closure: Let a, b ∈ G, then there exist integers m, n, p, q such that a = 2m 3n and b = 2 p 3q .
Then ab = (2m 3n )(2 p 3q ) = 2m+ p 3n+q . Thus, ab ∈ G.
• Associative: ∀ a, b, c ∈ G, ( ab)c = a(bc).
• Identity: 1 = 20 30 so 1 ∈ G.
• Inverse: ∀ a ∈ G, a = 2m 3n , the inverse of a is a−1 = 2−m 3−n . Thus, a−1 ∈ G.
1 a b
Problem 4. Let G be the set of all 3 × 3 real matrices of the form 0 1 c with a, b, c ∈ R. Then
0 0 1
G is a group under multiplication, as followed:
1 a b
• Closure: Let A, B ∈ G, then there exist real numbers a, b, c, r, s, t such that A = 0 1 c
0 0 1
1 r s
1 a + r b + at + s
1
c + t . Thus, AB ∈ G.
and B = 0 1 t . Then AB = 0
0 0 1
0
0
1
1
• Associative: ∀ A, B, C ∈ G, ( AB)C = A( BC ) (matrix multiplication is associative).
1 0 0
• Identity: The identity 3 × 3 matrix I3 = 0 1 0 with a = b = c = 0. So I3 ∈ G.
0 0 1
1 a b
1 − a ac − b
−c . Thus, A−1 ∈ G.
• Inverse: ∀ A ∈ G s.t. A = 0 1 c , then A−1 = 0 1
0 0 1
0 0
1
Problem 5. Let G be a finite group of even order. Let H = { a ∈ G | a−1 6= a}. First observe that
since e ∈
/ H, H is a proper subset of G, i.e., G \ H 6= ∅.
If a ∈ H then a−1 ∈ H, so H has an even number of elements. Now since G has even order,
G \ H also has an even number of elements. Thus, G \ H contains an element beside the identity.
Therefore, ∃ x ∈ G, x 6= e such that x = x −1 . This implies x2 = e.
Problem 6. Let G be a group and let g ∈ G. Suppose | g| = n, for some positive integer n. Then
( g−1 )n = ( gn )−1 = e−1 = e.
Now suppose that ( g−1 )k = e for some positive integer k. Using the same argument as above,
we get gk = e. Since | g| = n, n ≤ k.
Thus, n is the smallest positive integer such that ( g−1 )n = e. Hence, | g−1 | = n.
Problem 7. Recall from Chapter 1 that the 8 elements of D4 are: R0 , R90 , R180 , R270 (rotation
counter-clockwise of 0◦ , 90◦ , 180◦ , 270◦ , respectively), H (flip about the horizontal axis), V (flip
about the vertical axis), D and D 0 (flip about the diagonals).
D4 has three subgroups of order 4, namely:
{ R0 , R90 , R180 , R270 }, { R0 , R180 , H, V }, { R0 , R180 , D, D 0 }
Problem 8. Let G be an Abelian group and let g, h ∈ G. Suppose that both g and h have finite
order, then there exist positive integers m, n < ∞ such that | g| = m and |h| = n. Now since G is
Abelian, we have:
( gh)mn = gmn .hmn = ( gm )n .(hn )m = en .em = e
Hence, | gh| ≤ mn < ∞, and thus gh has finite order.
1 0
1 0
Problem 9. Let A =
and B =
be matrices in GL(2, R). Then | A| = | B| = 2
1 −1
2 −1
2
2
1 0
1 0
1 0
since
= I2 =
. Also, AB =
.
1 −1
2 −1
−1 1
2
1 0
Using induction, one can show that
=
for any positive integer n. This shows
−n 1
that the identity matrix I2 never appears as a positive power of AB. Thus, AB has infinite order.
( AB)n
3
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