*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 : **Z**_{4} -> **Z**_{10},
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 **Z**_{4}.

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*Solution:*
Example 3.7.5 shows that any group homomorphism from
**Z**_{18} into **Z**_{30}
must have the form

µ ( [x]_{18} ) = [mx]_{30},
for all [x]_{18} in **Z**_{18}. 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 **Z**_{30},
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 3^{2} 2 and
3^{3} 6,
it follows that [3] must have order 6.

(b) Show that **Z**_{17}^{×} is cyclic,
with generator [3]_{17}.

*Solution:*
The element [3] is a generator for **Z**_{17}^{×},
since

3^{2} = 9,

3^{3} = 27 10,

3^{4} 3 ·
10 13,

3^{5} 3 ·
13 5,

3^{6} 3 ·
5 15,

3^{7} 3 ·
15 11,

3^{8} 3 ·
11 16
1.

(c) Completely determine all group homomorphisms from
**Z**_{17}^{×} into
**Z**_{7}^{×}.

*Solution:*
Any group homomorphism
µ : **Z**_{17}^{×} ->
**Z**_{7}^{×}
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 **Z**_{17}^{×},
and in the second case

µ ( ([3]_{17})^{n} ) = [-1]^{n}_{7},
for all [x]_{17} = ([3]_{17})^{n}
in **Z**_{17}^{×}.

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

*Solution:*
If y_{1}
y_{2} (mod 6),
then 2y_{1} - 2y_{2} is divisible by 12, so

2y_{1}
2y_{2} (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
**Z**_{4} × **Z**_{6} onto
**Z**_{4} × **Z**_{3}.
Thus

(**Z**_{4} × **Z**_{6}) / { (0,0), (2,3) }
**Z**_{4} × **Z**_{3} .

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µ :

µ ( [x]_{n} ) = [x]_{m},
for all [x]_{n} in **Z**_{n}^{×},

*Solution:*
First, µ is a well-defined function, since if

[x_{1}]_{n} = [x_{2}]_{n}
belongs to **Z**_{n}^{×}, then

n | (x_{1} - x_{2}),
and this implies that

m | (x_{1} - x_{2}), since m | n. Thus

[x_{1}]_{m} = [x_{2}]_{m}, and so
µ ([x_{1}]_{n}) = µ ([x_{2}]_{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 **Z**_{36}^{×},

*Solution:*
Problem 21
shows that µ is a group homomorphism.
It is evident that µ maps **Z**_{36}^{×} onto
**Z**_{12}^{×},
since if gcd (x,12) = 1, then gcd (x,36) = 1.
The kernel of µ consists of the elements in
**Z**_{36}^{×}
that are congruent to 1 mod 12, namely

[1]_{36}, [13]_{36}, [25]_{36}.
It follows that

**Z**_{12}^{×}
**Z**_{36}^{×} / < [13]_{36} >.

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µ : G -> G

µ (x) = (µ_{1} (x), µ_{2} (x) ),
for all x in G,

*Solution:*
Given a, b in G,
we have

µ (a b)
= ( µ_{1} (ab), µ_{2} (ab) )

= ( µ_{1} (a) µ_{1} (b),
µ_{2} (a) µ_{2} (b))

µ (a) µ (b)
=( µ_{1} (a), µ_{2} (a)) ·
( µ_{1} (b), µ_{2} (b))

= ( µ_{1} (a) µ_{1} (b),
µ_{2} (a) µ_{2} (b))

and so µ : G -> G_{1} × G_{2}
is a group homomorphism.

Next problem | Next solution | Up | Table of Contents

*Solution:*
Using Problem 21, we can define group
homomorphisms

µ_{1} : **Z**_{pq}^{×} ->
**Z**_{p}^{×} and
µ_{2} : **Z**_{pq}^{×} ->
**Z**_{q}^{×} by setting

µ_{1} ( [x]_{pq} ) = [x]_{p},
for all [x]_{pq} in **Z**_{pq}^{×}, and

µ_{2} ( [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})
= (µ_{1} ([x]_{pq}),
µ_{2} ([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.

Forward to §3.8 | Forward to §3.8 problems | Back to §3.6 problems | Up | Table of Contents

Forward to §3.8 | Forward to §3.8 problems | Back to §3.6 problems | Up | Table of Contents