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

3^{2} = 9,

3^{3} = 27 10,

3^{4} 3 ·
10 30
13,

3^{5} 3 ·
13 39
5,

3^{6} 3 ·
5 15,

3^{7} 3 ·
15 45
11, and

3^{8} 3 ·
11 33
-1
1.

Therefore **Z**_{17}^{×}
is a cyclic group with 16 elements.
This provides the clue at to how to define
the isomorphism we need,
since **Z**_{16} is also a cyclic group,
with generator [1]_{16},
and Proposition 3.4.3(a) implies that
any isomorphism between cyclic groups
must map a generator to a generator.

Define
µ : **Z**_{16} -> **Z**_{17}^{×}
by setting

µ ([1]_{16}) = [3]_{17},

µ ([2]_{16}) = [3]_{17}^{2},

µ ([3]_{16}) = [3]_{17}^{3}, etc.

The general formula is

µ ([n]_{16}) = [3]_{17}^{n},
for all [n]_{16} in **Z**_{16}.

Since µ
is defined by using a representative n
of the equivalence class [n]_{16},
we have to show that the formula for µ
does not depend on the particular representative that is chosen.
If k m (mod 16),
then it follows from Proposition 3.2.8(c) that

[3]_{17}^{k} = [3]_{17}^{m}
since [3]_{17} has order 16 in **Z**_{17}^{×}

Therefore µ
([k]_{16}) =
µ
([m]_{16}),
and so µ
*is* a well-defined function.

Proposition 3.2.8(c) shows that
µ ([k]_{16}) = µ ([m]_{16})
only if k m (mod 16),
and so µ is a one-to-one function.
Then because both **Z**_{16} and
**Z**_{17}^{×}
have 16 elements,
it follows from Proposition 2.1.5
that µ is also an onto function.
The proof that µ respects the two group operations
follows the proof in Example 3.4.1.
For any elements [n]_{16} and [m]_{16}
in **Z**_{16},
we first compute what happens if we combine
[n]_{16} and [m]_{16}
using the operation in **Z**_{16},
and then substitute the result into the function
µ :

µ
([n]_{16} + [m]_{16})
= µ
([n + m]_{16} )
= [3]_{17}^{n+m} .

Next, we first apply the function µ to the two elements,
[n]_{16} and [m]_{16},
and then combine the results using the operation in
**Z**_{17}^{×}:

µ
([n]_{16}) · µ
([m]_{16})
= [3]_{17}^{n} [3]_{17}^{m}
= [3]_{17}^{n+m} .

Thus
µ ([n]_{16} + [m]_{16})
= µ ([n]_{16}) · µ ([m]_{16}),
and this completes the proof that µ is a group isomorphism.

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*Solution:*
The function µ preserves multiplication in **R**^{×}
since for all a,b in **R**^{×} we have

µ (ab)
= (ab)^{3} = a^{3} b^{3}
= µ (a) µ (b).

The function is one-to-one and onto since for each y in
**R**^{×}
the equation µ (x) = y has the unique solution
.

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Define µ : G

µ (x_{1}, x_{2})
= (µ_{1} (x_{1}), µ_{2} (x_{2})),
for all (x_{1}, x_{2}) in
G_{1} × G_{2}.

*Solution:*
If (y_{1},y_{2}) in H_{1} × H_{2},
then since µ_{1} is an isomorphism
there is a unique element x_{1} in G_{1}
with y_{1} = µ_{1} (x_{1}).
Similarly, since µ_{2} is an isomorphism
there is a unique element x_{2} in G_{2}
with y_{2} = µ_{2} (x_{2}).
Thus there is a unique element
(x_{1},x_{2}) in G_{1} × G_{2} such that
(y_{1},y_{2}) = µ (x_{1},x_{2}),
and so µ is one-to-one and onto.

Given (a_{1},a_{2}) and (b_{1},b_{2}) in
G_{1} × G_{2}, we have

µ ((a_{1},a_{2}) · (b_{1},b_{2}))
= µ ((a_{1} b_{1},a_{2} b_{2}))
= ( µ_{1} (a_{1} b_{1}),
µ_{2} (a_{2} b_{2}))
= ( µ_{1} (a_{1}) µ_{1} (b_{1}),
µ_{2} (a_{2}) µ_{2} (b_{2}))

and

µ ((a_{1},a_{2})) ·
µ ((b_{1},b_{2}))
= ( µ_{1} (a_{1}),
µ_{2} (a_{2})) ·
(µ_{1} (b_{1}), µ_{2} (b_{2}))
= ( µ_{1} (a_{1})
µ_{1} (b_{1}),
µ_{2} (a_{2})
µ_{2} (b_{2})) .

This shows that µ : G_{1} × G_{2}
-> H_{1} × H_{2}
is a group isomorphism.

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*Solution:*
You can check that **Z**_{7}^{×} is cyclic of order 6,
generated by [3]_{7},
and that **Z**_{11}^{×} is cyclic of order 10,
generated by [2]_{11}.
Just as in Problem 21, you can show that

µ_{1} : **Z**_{6} ->
**Z**_{7}^{×}
defined by
µ_{1} ([n]_{6}) = [3]_{7}^{n} and

µ_{2} : **Z**_{10} ->
**Z**_{11}^{×}
defined by µ_{2} ([m]_{10})
= [2]_{11}^{m}

are group isomorphisms.
It then follows from Problem 23 that

µ : **Z**_{6} × **Z**_{10} ->
**Z**_{7}^{×} ×
**Z**_{11}^{×}
defined by

µ (([n]_{6},[m]_{10}))
= ([3]_{7}^{n}, [2]_{11}^{m}),
for all [n]_{6} in **Z**_{6} and all
[m]_{10} in **Z**_{10},
is a group isomorphism.

Next problem | Next solution | Up | Table of Contents

µ ( [n]_{30}, [m]_{2} )
= ( [n]_{10}, [4n+3m]_{6} ),

*Solution:*
If ( [n]_{30}, [m]_{2} ) and
( [k]_{30}, [j]_{2} )
are equal elements of **Z**_{30} × **Z**_{2}, then

30 | n-k and 2 | m-j. It follows that 10 | n-k, and so
[n]_{10} = [k]_{10}. Furthermore,

30 | 4(n-k), so 6 | 4(n-k), and then 6 | 3(m-j),
which together imply that

6 | (4n+3m)-(4k+3j),
showing that [4n+3m]_{6} = [4k+3j]_{6}.
Thus

( [n]_{10}, [4n+3m]_{6} )
= ( [k]_{10}, [4k+3j]_{6} ),
which shows that

the formula for µ
*does* yield a well-defined function.

For any elements ([a]_{30},[c]_{2})
and ([b]_{30},[d]_{2}) we have

µ ( ([a]_{30},[c]_{2}) + ([b]_{30},[d]_{2}) )

= µ ( ([a+b]_{30},[c+d]_{2}) )

= ([a+b]_{10},[4(a+b) + 3(c+d)]_{2})

= ([a+b]_{10},[4a+4b+3c+3d]_{2})

µ ( ([a]_{30},[c]_{2}) )
+ µ ( ([b]_{30},[d]_{2}) )

= ( [a]_{10},[4a+3c]_{2})
+ ([b]_{10},[4b+3d]_{2})

= ([a+b]_{10},[4a+3c+4b+3d]_{2})

= ([a+b]_{10},[4a+4b+3c+3d]_{2})

and so µ respects the operations in the two groups.
This means that we can use Proposition 3.4.4
to show that µ is one-to-one.

If µ ( [n]_{30}, [m]_{2} )
= ( [0]_{10}, [0]_{6} ), then

( [n]_{10}, [4n+3m]_{6} )
= ( [0]_{10} , [0]_{6} ), so

10 | n, say n = 10q, for some q in **Z**, and

6 | (4n+3m), or 6 | (40q+3m). It follows that

2 | (40q+3m) and 3 | (40q+3m), and therefore

2 | 3m since 2 | 40q, and 3 | 40q since 3 | 3m.

Then since 2 and 3 are prime numbers, it follows that

2 | m, so [m]_{2} = [0]_{2}, and 3 | q, so

[n]_{30} = [10q]_{30} = [0]_{30}.
We have now shown that if

µ ( [n]_{30}, [m]_{2} )
= ( [0]_{10} , [0]_{6} ),
then

( [n]_{30}, [m]_{2} )
= ( [0]_{30}, [0]_{2} ),
and so the condition in Proposition
3.4.4 is satisfied.
We conclude that µ is a one-to-one function.
Since the two groups both have 60 elements,
it follows that µ must also be an onto function.
We have therefore checked all of the necessary conditions,
so we may conclude that µ is a group isomorphism.

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aHa^{-1} = { g in G | g = aha^{-1}
for some h in H }

*Solution:*
By Exercise 3.4.13 in the text, the function
µ : G -> G defined by µ (x) = axa^{-1},
for all x in G, is a group isomorphism.
By Exercise 3.4.15 the image under µ
of any subgroup of G is again a subgroup of G,
so aHa^{-1} = µ (H) is a subgroup of G.
It is then clear that the function
µ_{H} : H -> aHa^{-1} defined by
µ_{H} (x) = axa^{-1} is an isomorphism.

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(i) H K = { e },

(ii) HK = G, and

(iii) hk=kh for all h in H and k in K.

*Solution:*
Let µ
: G_{1} × G_{2} -> G be an isomorphism.
Exercise 3.3.9 in the text shows that in
G_{1} × G_{2} the subgroups

H^{*} = { (x_{1},x_{2}) | x_{2} = e } and
K^{*} = { (x_{1},x_{2}) | x_{1} = e }
have the properties we are looking for.

Let H = µ
(H^{*}) and K = µ
(K^{*}) be the images in G
of H^{*} and K^{*}, respectively.
We know (by Exercise 3.4.15) that H and K are subgroups of G,
so we only need to show that

H K = { e }, HK = G,
and hk=kh for all h in H and k in K.

Let y be in G, with y = µ (x),
for x in G_{1} × G_{2}.
If y is in H K,
then y is in H, and so x is in H^{*}.
Since y is in K as well, we must also have x in K^{*}, so x is in
H^{*} K^{*},
and therefore x = (e_{1},e_{2}),
where e_{1} and e_{2} are the respective identity elements
in G_{1} and G_{2}.
Thus y = µ ((e_{1},e_{2})) = e,
showing that H K = { e }.
Since y is any element of G,
and we can write x = h^{*} k^{*}
for some h^{*} in H^{*}
and some k^{*} in K^{*},
it follows that y = µ
(h^{*} k^{*}) = µ
(h^{*}) µ
(k^{*}),
and thus G = HK.
It is clear that µ
preserves the fact that elements of
H^{*} and K^{*} commute.
We conclude that H and K satisfy the desired conditions.

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*Solution:*

Since each matrix in GL_{2} (**Z**_{p} )
has nonzero determinant,
it is clear that the mapping

µ : **Z**_{p}^{×} ×
**Z**_{p}^{×} ->
GL_{2} (**Z**_{p} ) defined by

µ (x_{1},x_{2}) =

for each (x_{1},x_{2}) in
**Z**_{p}^{×} ×
**Z**_{p}^{×},
is one-to-one and maps
**Z**_{p}^{×} ×
**Z**_{p}^{×}
onto the subgroup of diagonal matrices.
This mapping respects the operations in the two groups, since for

(a_{1}, a_{2}), (b_{1}, b_{2}) in
**Z**_{p}^{×} ×
**Z**_{p}^{×}
we have

µ ((a_{1},a_{2}) (b_{1},b_{2}))
= µ ((a_{1} b_{1},a_{2} b_{2}))

= =

= µ ((a_{1},a_{2})) µ
((b_{1},b_{2})) .

Thus µ is the desired isomorphism.

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H =

is a subgroup of G.*Solution:*
Closure:
The general product
=

has the correct form to belong to H.

Identity: The identity matrix has the correct form.

Existence of inverses:
The inverse
=

of any element in H has the correct form to belong H.

(b) Show that H is isomorphic
to the group **R** of all real numbers, under addition.

*Solution:*
Define µ : **R** -> H by

µ (x) =
,
for all x in **R**.

You can easily check that µ is an isomorphism.
(The computation necessary to show that µ
preserves the respective operations
is the same computation we used to show that H is closed.)

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G = .

Show that G is not isomorphic to the direct product

*Solution:*
Our approach is to try to find an algebraic property
that would be preserved by any isomorphism
but which is satisfied by only one of the two groups in question.
By Proposition 3.4.3 (b),
if one of the groups is abelian but the other is not,
then the groups cannot be isomorphic.

By Exercise 3.3.5 in the text, the direct product
**R**^{×} × **R**
is an abelian group, since each factor is abelian.
On the other hand, G is not abelian, since

=

but

= .

Thus the two groups cannot be isomorphic.

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H =

Show that H is isomorphic to the symmetric group S*Solution:*
This group is small enough that we can just compare its multiplication table
to that of S_{3},
as given in Table 3.3.3 of the text
(see page 104 of the text or the
online table).
Remember that constructing an isomorphism
is the same as constructing a one-to-one correspondence
between the elements of the group,
such that all entries in the respective group tables
also have the same one-to-one correspondence.

In this case we can explain how this can be done, without actually writing out the multiplication table. Let

A = and B = .

Then just as in Problem 3.3.24,
we can show that BA = A^{-1} B,
and that each element of H has the form
can be written uniquely in the form

A^{i} B^{j},
where i = 0,1,2,3 or 4 and j = 0 or 1.

This information should make it plausible that the function
µ : S_{3} -> H
defined by
µ (a^{i} b^{j})
= A^{i} B^{j},
for i = 0,1,2,3 or 4 and j = 0 or 1,
gives a one-to-one correspondence between the elements of the groups
which also produces multiplication tables that look exactly the same.

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*Solution:*
**(Outline only)**
The operation is well-defined on S, since µ and
µ ^{-1} are functions
and the operation on G is well-defined.
The associative law holds in S because it holds in G;
the identity element in S is µ (e),
where e is the identity of G,
and it is easy to check that if x in S, then

x^{-1} = µ ( (µ ^{-1} (x))^{-1} ).

*Comment:*
This reveals the secret behind problems like
Exercises 3.1.11 and 3.4.12 in the text.
Given a known group G such as **R**^{×},
we can use one-to-one functions defined on G
to produce new groups with operations
that look rather different from the usual examples.

Forward to §3.5 | Forward to §3.5 problems | Back to §3.3 problems | Up | Table of Contents

Forward to §3.5 | Forward to §3.5 problems | Back to §3.3 problems | Up | Table of Contents