§ 3.4 Isomorphisms: Solved problems

21. Show that Z17× is isomorphic to Z16.

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

32 = 9,
33 = 27 10,
34 3 · 10 30 13,
35 3 · 13 39 5,
36 3 · 5 15,
37 3 · 15 45 11, and
38 3 · 11 33 -1 1.

Therefore Z17× is a cyclic group with 16 elements. This provides the clue at to how to define the isomorphism we need, since Z16 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 : Z16 -> Z17× by setting

µ ([1]16) = [3]17,
µ ([2]16) = [3]172,
µ ([3]16) = [3]173, etc.

The general formula is

µ ([n]16) = [3]17n, for all [n]16 in Z16.

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]17k = [3]17m since [3]17 has order 16 in Z17×
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 Z16 and Z17× 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 Z16, we first compute what happens if we combine [n]16 and [m]16 using the operation in Z16, and then substitute the result into the function µ :
µ ([n]16 + [m]16) = µ ([n + m]16 ) = [3]17n+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 Z17×:
µ ([n]16) · µ ([m]16) = [3]17n [3]17m = [3]17n+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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22. Let µ : R× -> R× be defined by µ (x) = x3, for all x in R. Show that µ is a group isomorphism.

Solution: The function µ preserves multiplication in R× since for all a,b in R× we have
µ (ab) = (ab)3 = a3 b3 = 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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23. Let G1, G2, H1, H2 be groups, and suppose that µ1 : G1 -> H1 and µ2 : G2 -> H2 are group isomorphisms.
Define µ : G1 × G2 -> H1 × H2 by

µ (x1, x2) = (µ1 (x1), µ2 (x2)),     for all (x1, x2) in G1 × G2.

Prove that µ is a group isomorphism.

Solution: If (y1,y2) in H1 × H2, then since µ1 is an isomorphism there is a unique element x1 in G1 with y1 = µ1 (x1). Similarly, since µ2 is an isomorphism there is a unique element x2 in G2 with y2 = µ2 (x2). Thus there is a unique element (x1,x2) in G1 × G2 such that (y1,y2) = µ (x1,x2), and so µ is one-to-one and onto.

Given (a1,a2) and (b1,b2) in G1 × G2, we have

µ ((a1,a2) · (b1,b2)) = µ ((a1 b1,a2 b2)) = ( µ1 (a1 b1), µ2 (a2 b2)) = ( µ1 (a1) µ1 (b1), µ2 (a2) µ2 (b2))
and
µ ((a1,a2)) · µ ((b1,b2)) = ( µ1 (a1), µ2 (a2)) · (µ1 (b1), µ2 (b2)) = ( µ1 (a1) µ1 (b1), µ2 (a2) µ2 (b2)) .

This shows that µ : G1 × G2 -> H1 × H2 is a group isomorphism.

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24. Prove that the group Z7× × Z11× is isomorphic to the group Z6 × Z10.

Solution: You can check that Z7× is cyclic of order 6, generated by [3]7, and that Z11× is cyclic of order 10, generated by [2]11. Just as in Problem 21, you can show that
µ1 : Z6 -> Z7× defined by µ1 ([n]6) = [3]7n and
µ2 : Z10 -> Z11× defined by µ2 ([m]10) = [2]11m
are group isomorphisms. It then follows from Problem 23 that
µ : Z6 × Z10 -> Z7× × Z11× defined by
µ (([n]6,[m]10)) = ([3]7n, [2]11m), for all [n]6 in Z6 and all [m]10 in Z10, is a group isomorphism.

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25. Define µ : Z30 × Z2 -> Z10 × Z6 by

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

for all ( [n]30, [m]2 ) in Z30 × Z2 . First prove that µ is a well-defined function, and then prove that µ is a group isomorphism.

Solution: If ( [n]30, [m]2 ) and ( [k]30, [j]2 ) are equal elements of Z30 × Z2, 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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26. Let G be a group, and let H be a subgroup of G. Prove that if a is any element of G, then the subset

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

is a subgroup of G that is isomorphic to 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 µH : H -> aHa-1 defined by µH (x) = axa-1 is an isomorphism.

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27. Let G, G1, G2 be groups. Prove that if G is isomorphic to G1 × G2, then there are subgroups H and K in G such that
(i) H K = { e },
(ii) HK = G, and
(iii) hk=kh for all h in H and k in K.

Solution: Let µ : G1 × G2 -> G be an isomorphism. Exercise 3.3.9 in the text shows that in G1 × G2 the subgroups
H* = { (x1,x2) | x2 = e } and K* = { (x1,x2) | x1 = 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 G1 × G2. 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 = (e1,e2), where e1 and e2 are the respective identity elements in G1 and G2. Thus y = µ ((e1,e2)) = 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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28. Show that for any prime number p, the subgroup of diagonal matrices in GL2 (Zp ) is isomorphic to Zp× × Zp×.

Solution:

Since each matrix in GL2 (Zp ) has nonzero determinant, it is clear that the mapping
µ : Zp× × Zp× -> GL2 (Zp ) defined by

µ (x1,x2) =

for each (x1,x2) in Zp× × Zp×, is one-to-one and maps Zp× × Zp× onto the subgroup of diagonal matrices. This mapping respects the operations in the two groups, since for
(a1, a2), (b1, b2) in Zp× × Zp× we have

µ ((a1,a2) (b1,b2)) = µ ((a1 b1,a2 b2))

= =

= µ ((a1,a2)) µ ((b1,b2)) .

Thus µ is the desired isomorphism.

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29. (a) In the group G = GL2 (R) of invertible 2 × 2 matrices with real entries, show that

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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30. Let G be the subgroup of \GL2 (R) defined by
G = .
Show that G is not isomorphic to the direct product R× × R.

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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31. Let H be the following subgroup of the group G = GL2 (Z3).

H =

Show that H is isomorphic to the symmetric group S3.

Solution: This group is small enough that we can just compare its multiplication table to that of S3, 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

Ai Bj, where i = 0,1,2,3 or 4 and j = 0 or 1.

This information should make it plausible that the function µ : S3 -> H defined by µ (ai bj) = Ai Bj, 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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32. Let G be a group, and let S be any set for which there exists a one-to-one and onto function µ : G -> S. Define an operation on S by setting x1 · x2 = µ ( µ -1 (x1) µ -1 (x2) ), for all x1, x2 in S. Prove that S is a group under this operation, and that µ is actually a group isomorphism.

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.

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