MATH 433
May 1, 2015
Exam 3: Solutions
Problem 1 (20 pts.) Let G and H be groups. Suppose g ∈ G is an element of
order 6 and h ∈ H is an element of order 4. Find the order of the element (g, h) of the
group G × H.
Solution: 12.
By definition of the direct product G × H, we have (g, h)n = (gn , hn ) for any integer n > 0.
Furthermore, the identity element of the group G × H is (eG , eH ), where eG is the identity
element of G and eH is the identity element of H. Therefore (g, h)n = (eG , eH ) if and only if
g n = eG and hn = eH . The condition gn = eG holds if n is a multiple of o(g) = 6. The condition
hn = eH holds if n is a multiple of o(h) = 4. Both conditions hold simultaneously if n is a
common multiple of 6 and 4. The least number with this property is lcm(6, 4) = 12.
Problem 2 (20 pts.) Find all subgroups of the group G11 (that is, multiplicative
group of invertible congruence classes modulo 11).
Solution: {[1]}, {[1], [10]}, {[1], [3], [4], [5], [9]}, and G11 .
The group G11 has 10 elements as all nonzero congruence classes modulo 11 are invertible.
The element g = [2] generates the entire group as g 2 = [4], g 3 = [8], g4 = [16] = [5], g 5 = [32] =
[10] = [−1], g 6 = [−2] = [9], g 7 = [−4] = [7], g 8 = [−8] = [3], g 9 = [−16] = [6], and g 10 = [1].
Therefore G11 is a cyclic group and so are all subgroups of G11 . Any finite cyclic group has a
unique subgroup of order n for every n dividing the order of the group. In our case, the group
G11 , in addition to the trivial subgroup and G11 itself, has two other subgroups: {g 0 , g5 } and
{g 0 , g2 , g4 , g6 , g8 }.
Problem 3 (20 pts.)
Consider cyclic groups Z20 , Z25 and Z30 . Suppose that
elements g1 ∈ Z20 , g2 ∈ Z25 and g3 ∈ Z30 are chosen so that the set of their orders is
{5, 6, 10}. Determine the order for each of the elements g1 , g2 and g3 .
Solution: o(g1 ) = 10, o(g2 ) = 5, and o(g3 ) = 6.
As follows from Lagrange’s Theorem, the order of a finite group is divisible by orders of all
elements of the group. Therefore o(g1 ) is a divisor of 20, o(g2 ) is a divisor of 25, and o(g3 ) is a
divisor of 30.
Since 20 and 25 are not divisible by 6, it follows that 6 is the order of g3 . Since 25 is not
divisible by 10 and the order of g3 is already determined to be 6, we conclude that 10 is the
order of g1 . The remaining order 5 must be that of g2 .
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Problem 4 (20 pts.)
matrix
Let f be a linear coding function defined by the generator


1 0 0 1 1 0 1
0 1 0 1 0 1 1 .
0 0 1 0 1 1 1
Suppose that a message encoded by f is received with errors as 0110111 1100010 1000000.
Correct errors and decode the received message.
Solution: 001 110 000.
The coding function f acts from B3 to B7 , where B = {0, 1}. It is given by f (w) = wG,
where G is the generator matrix, w ∈ B3 is regarded as a row vector, and the arithmetic is done
modulo 2. The 8 codewords are linear combinations of rows of the generator matrix:
0000000 0010111 0101011 1001101 0111100 1011010 1100110 1110001.
Every received word is corrected to the closest codeword. The corrected message is
0010111 1100110 0000000.
The code is systematic since the first 3 columns of the generator matrix form the identity matrix.
Therefore decoding consists of truncating the codewords to 3 digits: 001 110 000.
Problem 5 (20 pts.) Assume that a polynomial p(x) = x3 − 3x2 − 14x + 20 over R
is a greatest common divisor of two non-zero polynomials f, g ∈ R[x]. Find all common
roots of the polynomials f and g.
√
√
Solution: 5, −1 + 5, and −1 − 5.
A real number α is a common root of the polynomials f and g if and only if it is a root of
their greatest common divisor p. Indeed, if f (α) = g(α) = 0, then f and g are both divisible by
x − α. By definition of the greatest common divisor, the polynomial p is also divisible by x − α
so that p(α) = 0. Conversely, if p(α) = 0 then p is divisible by x − α. Since p is a divisor of f
and of g, it follows that f (α) = g(α) = 0.
Since p is a polynomial with integer coefficients, its integer root is a divisor of the constant
term 20. Evaluating p at divisors of 20, we eventually find that p(5) = 0. Dividing p by x − 5, we
obtain: x3 − 3x2 − 14x + 20 = (x − 5)(x2 + 2x − 4). Hence the remaining roots of the polynomial
p (if any) are solutions of the equation
√
x2 + 2x − 4 = 0 ⇐⇒ x2 + 2x + 1 = 5 ⇐⇒ (x + 1)2 = 5 ⇐⇒ x = −1 ± 5.
Bonus Problem 6 (15 pts.)
are not isomorphic to each other.
Find two non-commutative groups of order 12 that
Solution: S(3) × Z2 and A(4).
The groups S(3) × Z2 and A(4) are both non-commutative and of order 12. To prove that
these groups are not isomorphic, we shall show that one of them contains an element of order 6
while the other does not. Indeed, an element ((1 2 3), [1]2 ) of the group S(3) × Z2 has order 6.
On the other hand, the group A(4) contains permutations of the following cycle shapes: (1 2 3),
(1 2)(3 4), and id. Neither of them has order 6.
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