§ 3.7 Homomorphisms: Solved problems

Solution: Example 3.7.5 shows that any group homomorphism from Z_n into Z_k must have the form
µ ( [x]_n ) = [mx]_k, for all [x]_n in Z_n. Under any group homomorphism
µ : Z₄ -> Z₁₀, the order of µ ([1]₄) must be a divisor of 4 and of 10, so the only possibilities are 1 and 2. Thus
µ ([1]₄) = [0]₁₀, which defines the zero function, or else

µ ([1]₄) = [5]₁₀, which leads to the formula µ ([x]₄) = [5x]₁₀, for all [x]₄ in Z₄.

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Solution: Example 3.7.5 shows that any group homomorphism from Z₁₈ into Z₃₀ must have the form
µ ( [x]₁₈ ) = [mx]₃₀, for all [x]₁₈ in Z₁₈. Since
gcd (18,30) = 6, the possible orders of
[m]₃₀ = µ ([1]₁₈) are 1,2,3,6.
The corresponding choices for [m]₃₀ are
[0]₃₀, of order 1,
[15]₃₀, of order 2,
[10]₃₀ and [20]₃₀, of order 3, and
[5]₃₀ and [25]₃₀, of order 6.

(b) Choose one of the nonzero formulas in part (a), and for this formula find the kernel and image, and show how elements of the image correspond to cosets of the kernel.

Solution: For example, consider µ ([x]₁₈) = [5x]₃₀. The image of µ consists of the multiples of 5 in Z₃₀, which are
0,5,10,15,20,25. We have ker (µ) = {0,6,12}, and then cosets of the kernel are defined by adding
1, 2, 3, 4, and 5, respectively. We have the following correspondence

{ 0,6,12 }   <--> µ (0) = 0,
{ 3,9,15 }   <--> µ (3) = 15,
{ 1,7,13 }   <--> µ (1) = 5,
{ 4,10,16 } <--> µ (4) = 20,
{ 2,8,14 }   <--> µ (2) = 10,
{ 5,11,17 } <--> µ (5) = 25.

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Solution: Since 3² 2 and 3³ 6, it follows that [3] must have order 6.

(b) Show that Z₁₇^× is cyclic, with generator [3]₁₇.

Solution: The element [3] is a generator for Z₁₇^×, since

3² = 9,
3³ = 27 10,
3⁴ 3 · 10 13,
3⁵ 3 · 13 5,
3⁶ 3 · 5 15,
3⁷ 3 · 15 11,
3⁸ 3 · 11 16 1.

(c) Completely determine all group homomorphisms from Z₁₇^× into Z₇^×.

Solution: Any group homomorphism µ : Z₁₇^× -> Z₇^× is determined by its value on the generator [3]₁₇, and the order of µ([3]₁₇) must be a common divisor of 16 and 6. The only possible orders are 1 and 2, so either
µ ( [3]₁₇ ) = [1]₇ or µ ( [3]₁₇ ) = [-1]₇. In the first case,
µ ( [x]₁₇ ) = [1]₇ for all [x]₁₇ in Z₁₇^×, and in the second case
µ ( ([3]₁₇)ⁿ ) = [-1]ⁿ₇, for all [x]₁₇ = ([3]₁₇)ⁿ in Z₁₇^×.

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(a) Show that µ is a well-defined group homomorphism.

Solution: If y₁ y₂ (mod 6), then 2y₁ - 2y₂ is divisible by 12, so
2y₁ 2y₂ (mod 4), and then it follows quickly that µ is a well-defined function. It is also easy to check that µ preserves addition.

(b) Find the kernel and image of µ, and apply the fundamental homomorphism theorem.

Solution: If (x,y) belongs to ker (µ), then y 0 (mod 3), so y = 0 or y = 3. If y = 0, then x = 0, and if y = 3, then x = 2. Thus the elements of the kernel K are (0,0) and (2,3). It follows that there are 24 / 2 = 12 cosets of the kernel. These cosets are in one-to-one correspondence with the elements of the image, so µ must map Z₄ × Z₆ onto Z₄ × Z₃. Thus

(Z₄ × Z₆) / { (0,0), (2,3) } Z₄ × Z₃ .

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µ ( [x]_n ) = [x]_m,     for all [x]_n in Z_n^×,

Solution: First, µ is a well-defined function, since if
[x₁]_n = [x₂]_n belongs to Z_n^×, then
n | (x₁ - x₂), and this implies that
m | (x₁ - x₂), since m | n. Thus
[x₁]_m = [x₂]_m, and so µ ([x₁]_n) = µ ([x₂]_n).

Next, µ is a homomorphism since for [a]_n, [b]_n in Z_n^×,

µ ([a]_n [b]_n) = µ ([ab]_n) = [ab]_m = [a]_m [b]_m = µ ([a]_n) µ ([b]_n) .

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µ ( [x]₃₆ ) = [x]₁₂,     for all [x]₃₆ in Z₃₆^×,

Solution: Problem 21 shows that µ is a group homomorphism. It is evident that µ maps Z₃₆^× onto Z₁₂^×, since if gcd (x,12) = 1, then gcd (x,36) = 1. The kernel of µ consists of the elements in Z₃₆^× that are congruent to 1 mod 12, namely
[1]₃₆, [13]₃₆, [25]₃₆. It follows that
Z₁₂^× Z₃₆^× / < [13]₃₆ >.

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µ (x) = (µ₁ (x), µ₂ (x) ),     for all x in G,

Solution: Given a, b in G, we have

µ (a b) = ( µ₁ (ab), µ₂ (ab) )
= ( µ₁ (a) µ₁ (b), µ₂ (a) µ₂ (b))

µ (a) µ (b) =( µ₁ (a), µ₂ (a)) · ( µ₁ (b), µ₂ (b))
= ( µ₁ (a) µ₁ (b), µ₂ (a) µ₂ (b))

and so µ : G -> G₁ × G₂ is a group homomorphism.

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Solution: Using Problem 21, we can define group homomorphisms
µ₁ : Z_pq^× -> Z_p^× and µ₂ : Z_pq^× -> Z_q^× by setting
µ₁ ( [x]_pq ) = [x]_p, for all [x]_pq in Z_pq^×, and
µ₂ ( [x]_pq ) = [x]_q, for all [x]_pq in Z_pq^×.

Using Problem 23, we can define a group homomorphism
µ : Z_pq^× -> Z_p^× × Z_q^× by setting
µ ([x]_pq) = (µ₁ ([x]_pq), µ₂ ([x]_pq) ), for all [x]_pq in Z_pq^×.
If [x]_pq in ker (µ), then [x]_p = [1]_p and [x]_q = [1]_q,
so p | (x-1) and q | (x-1), and this implies that pq | (x-1),
since p and q are relatively prime. It follows that
[x]_pq = [1]_pq, and this shows that µ is a one-to-one function.
Exercise 1.4.27 in the text states that if
m > 0 and n > 0 are relatively prime integers, then
(mn) = (m) (n). It follows that
Z_pq^× and Z_p^× × Z_q^× have the same order, so µ is also an onto function.
This completes the proof that µ is a group isomorphism.

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