*Solution:*
First, we have
| **Z**_{9}^{×} | = 6, and
| **Z**_{18}^{×} | = 6.

In **Z**_{9}^{×}, we have

2^{2}=4,

2^{3}=8 1,

and so [2] must have order 6, showing that

**Z**_{9}^{×} is cyclic of order 6.
Our theorems tell us that

**Z**_{9}^{×}
**Z**_{6}.
In **Z**_{18}^{×},

5^{2} 7,

5^{3} 17
1,

and so [5] must have order 6, showing that

**Z**_{18}^{×} is cyclic of order 6.
Our theorems tell us that

**Z**_{18}^{×}
**Z**_{6}.
Thus all three groups are isomorphic.

Next problem | Next solution | Up | Table of Contents

*Solution:*
It follows from Theorem 3.5.4 that

**Z**_{10}
**Z**_{2} × **Z**_{5}, and that
**Z**_{20}
**Z**_{4} × **Z**_{5}.

It then follows from Problem 3.4.23 that

**Z**_{4} × **Z**_{10}
**Z**_{4} × **Z**_{2} × **Z**_{5},
and **Z**_{2} × **Z**_{20}
**Z**_{2} × **Z**_{4} × **Z**_{5}.
Finally,

it is possible to show that the obvious mapping from

**Z**_{4} × **Z**_{2} × **Z**_{5}
onto **Z**_{2} × **Z**_{4} ×
**Z**_{5}
is an isomorphism. Therefore

**Z**_{4} × **Z**_{10}
**Z**_{2} × **Z**_{20}.

Next problem | Next solution | Up | Table of Contents

*Solution:*
As in Problem 21,
**Z**_{4} × **Z**_{15}
**Z**_{4} × **Z**_{3} × **Z**_{5},
and

**Z**_{6} × **Z**_{10}
**Z**_{2} × **Z**_{3} ×
**Z**_{2} × **Z**_{5}.
The two groups are not isomorphic

since the first has an element of order 4,
while the second does not.

Next problem | Next solution | Up | Table of Contents

*Solution:*
The subgroups correspond to the divisors of 100,
and are given below:

Note that n **Z**_{100} is used to mean all multiples of n in **Z**_{100}.

Next problem | Next solution | Up | Table of Contents

*Solution:*
By Proposition 3.5.3(b),
the generators correspond to the numbers less than 28
and relatively prime to 28.
The
Euler
-function
allows us to
compute
how many there are:

(28) = (1/2) · (6/7) · 28 = 12.

The list of generators is
{ ± 1, ± 3, ± 5, ± 9, ± 11, ± 13 }.

Next problem | Next solution | Up | Table of Contents

*Solution:*
Using Proposition 3.5.3(a), we first find that

gcd (18,30) = 6.
Then < [18]_{30} > = < [6]_{30} >,

and so the subgroup has 30 / 6 = 5 elements.

Similarly, < [24]_{30} > = < [6]_{30} >,
and so we actually have

< [24]_{30} > = < [18]_{30} >.

Next problem | Next solution | Up | Table of Contents

*Solution:*
Since 7 and 11 are primes, the groups are cyclic.
If a has order 7 in G_{1} and b has order 11 in G_{2},
then (a,b) has order lcm [7,11] = 77 in G_{1} × G_{2}.
Thus G_{1} × G_{2} is cyclic since it has an element
whose order is equal to the order of the group.

Next problem | Next solution | Up | Table of Contents

*Solution:*
If G is cyclic of order 2n, for some positive integer n,
then it follows from Theorem 3.5.2
that G is isomorphic to **Z**_{2n}.
Since isomorphisms preserve orders of elements,
we only need to answer the question in **Z**_{2n}.
In that group, the elements of order 2 are the nonzero solutions
to the congruence 2x 0 (mod 2n),
and since the congruence can be rewritten as x 0 (mod n),
we see that [n]_{2n} is the only element of order 2 in **Z**_{2n}.

Next problem | Next solution | Up | Table of Contents

*Solution:*
In **Z**_{15}^{×}, both [-1]_{15} and [4]_{15}
are easily checked to have order 2.
In **Z**_{21}^{×}, we have [8]_{21}^{2} = [64]_{21} = [1]_{21},
and so [8]_{21} and [-1]_{21} have order 2.

Next problem | Next solution | Up | Table of Contents

*Solution:*
The quaternion group
Q = { ± 1, ± **i**, ± **j**, ± **k** }
is defined in Example 3.3.7 of the text
(see page 108 or the online table).
The elements satisfy the following identities:

**i**^{2} = **j**^{2} = **k**^{2} = -1

and

**i** **j** = **k**,
**j** **k** = **i**,
**k** **i** = **j**,
**j** **i** =-**k**,
**k** **j** =-**i**,
**i** **k** =-**j**.

The cyclic subgroups

< -1 > = { ± 1 },

< ± **i** > = { ± 1, ± **i** },

< ± **j** > = { ± 1, ± **j** }, and

< ± **k** > = { ± 1, ± **k** }

can be found by using the given identities.
For example,

**i**^{2} = -1,
**i**^{3} = **i**^{2} **i** = - **i**, and
**i**^{4} = **i**^{2} **i**^{2} = (-1)^{2} = 1.

In S_{4}, let (1,2,3,4) = a and (1,3) = b.

Since a is a cycle of length 4, it has order 4, with

a^{2} = (1,3)(2,4) and a^{3} = a^{-1} = (1,4,3,2).

To find the subgroup generated by a and b,
we have

ab = (1,2,3,4)(1,3) = (1,4)(2,3),

a^{2}b = (1,3)(2,4)(1,3) = (2,4), and

a^{3}b = (1,4,3,2)(1,3) = (1,2)(3,4).

On the other side, we have

ba = (1,3)(1,2,3,4) = (1,2)(3,4) = a^{3} b,

ba^{2} = (1,3)(1,3)(2,4) = (2,4) = a^{2} b, and

ba^{3} = (1,3)(1,4,3,2) = (1,4)(2,3) = ab.

This shows that the subgroup generated by a and b consists
of the 8 elements

{ e, a, a^{2}, a^{3}, b, ab, a^{2} b, a^{3} b }.

Furthermore, from the cycle structures of the elements
we can see that the only cyclic subgroup of order 4
is the one generated by a (and a^{3}).
In any isomorphism, cyclic subgroups would correspond to cyclic subgroups,
and so it is impossible for this group
to be isomorphic to the quaternion group,
which has 3 cyclic subgroups of order 4.

Next problem | Next solution | Up | Table of Contents

*Solution:*
We know that [-1]_{pq} has order 2,
so by Problem 27
it is enough to find one other element of order 2.
The Chinese remainder theorem
(Theorem 1.3.6)
states that the system of congruences

x 1 (mod p) and
x -1 (mod q)

has a solution [a]_{pq}, since p and q are relatively prime.
Because q is an odd prime,

[-1]_{pq} is not a solution, so
[a]_{pq}
[-1]_{pq}.
But

a^{2} 1 (mod p) and
a^{2} 1 (mod q),
so a^{2}
1 (mod pq) since p and q are relatively prime,
and thus [a]_{pq} has order 2.

Forward to §3.6 | Forward to §3.6 problems | Back to §3.4 problems | Up | Table of Contents

Forward to §3.6 | Forward to §3.6 problems | Back to §3.4 problems | Up | Table of Contents