Assignment 6c2, 3, 11, 13, 14, 22, 26, 35, 55, 56

2. Find Aut(Z).

Since Z = 〈1〉, by property 4 of Theorem 6.2 we know that any automorphism of Z must map1 to another generator of Z. Also, since automorphisms preserve the operation, we see that forany automorphism φ, φ(k) = kφ(1) (as in example 13 in the text). Since the only generators ofZ are 1,−1, that gives us two possibilities for automorphisms in Aut(Z), namely f(x) = x andg(x) = −x.

The identity function f(x) = x is clearly an automorphism (and should have been proven to bean isomorphism in problem 6 – showing that isomorphism is an equivalence relation).

We now show that g(x) = −x is an automorphism of Z. If x, y ∈ Z and g(x) = g(y) then−x = −y and thus x = y so g is 1-1. If z ∈ Z then so is −z and g(−z) = −(−z) = z so g isonto. If x, y ∈ Z then g(x + y) = −(x + y) = −x − y = −x + (−y) = g(x) + g(y) so g is OP.Therefore g is an automorphism of Z.

Since we have shown that these are the only two possibilities and that both actually are auto-morphisms, we see that Aut(Z) = {f, g} where f(x) = x and g(x) = −x.

3. Let R+ be the group of positive real numbers under multiplication. Show that the mappingφ(x) =

√x is an automorphism of R+.

Let x, y ∈ R+ and suppose that φ(x) = φ(y). Then√x =

√y and squaring both sides gives

x = y so φ is 1-1. Let z ∈ R+. Then z2 ∈ R+ and φ(z2) =√z2 = z (since z > 0) so φ is onto. If

x, y ∈ R+ then φ(xy) =√xy =

√x√y = φ(x)φ(y) so φ is OP. Therefore φ is an automorphism

of R+.

11. If g and h are elements from a group, prove that φgφh = φgh.

Recall that φa is the inner automorphism of G induced by a, that is: φa(x) = axa−1. To showthat the given functions are equal, I will show that for any x ∈ G they have the same outputvalue. Let x ∈ G. Then φgφh(x) = φg

(φh(x)

)= φg(hxh

−1) = g(hxh−1)g−1 = (gh)x(h−1g−1) =(gh)x(gh)−1 = φgh(x). Since x was an arbitrary element of G, we see that φgφh = φgh.

13. Prove the assertion in Example 12 that the inner automorphisms φR0 , φR90 , φH , and φD of D4

are distinct.

We need to show that for any pair of these functions, there is some element of D4 that givesdifferent outputs. First we’ll compare φR0 to all the others. Note that φR0(x) = R0xR

−10 = x

since R0 is the identity. Since φR0(H) = H and φR90(H) = R90HR270 = D′R270 = V we see thatφR0 6= φR90 . Since φR0(R90) = R90 and φH(R90) = HR90H = DH = R270 we see that φR0 6= φH .Also, since φD(R90) = DR90D = V D = R270 we see that φR0 6= φD. Thus φR0 is distinct fromthe other three inner automorphisms listed.

Now compare φR90 to the remaining two automorphisms. Since φR90(R90) = R90R90R−190 = R90

and φH(R90) = HR90H = DH = R270 and φD(R90) = DR90D = V D = R270 we see thatφR90 6= φH and φR90 6= φD.

It remains to show that φH 6= φD. Since φH(H) = HHH−1 = H and φD(H) = DHD = R270D =V we see that φH 6= φD.

Therefore, the four inner automorphisms φR0 , φR90 , φH , and φD of D4 are distinct.Note that there are probably lots of other choices that we could have used to see that thesefunctions are distinct, but we just needed one.

14. Find Aut(Z6).

Similar to problem 2, since Z6 is cyclic, we can only send a generator to another generator, andthen we need to check if those functions give automorphisms. Recall from an earlier chapter thatthe generators for Z6 will be those numbers relatively prime to 6, namely 1 and 5.

Consider f(x) = 1 · x = x. This is the identity function which we already know is always anautomorphism.

Now consider g(x) = 5·x. If x, y ∈ Z6 and g(x) = g(y) then 5x ≡ 5y mod 6. Then 5(5x) ≡ 5(5y)or 25x ≡ 25y which implies that x ≡ y mod 6 since 25 ≡ 1. Thus g is 1-1. Let z ∈ Z6. Then 5zmod 6 ∈ Z6 and g(5z) ≡ 25z ≡ z mod 6 so g is onto. If x, y ∈ Z6 then g(x + y) ≡ 5(x + y) ≡5x+ 5y ≡ g(x) + g(y) so g is OP. Thus g(x) = 5x is an automorphism.

Therefore Aut(Z6) = {f, g} where f(x) = x and g(x) = 5x.

22. Let φ be an automorphism of a group G. Prove that H = {x ∈ G | φ(x) = x} is a subgroup of G.

We will use the 2 Step Subgroup Test. Since for any automorphism we have φ(e) = e, we seethat e ∈ H so H 6= ∅. Let a, b ∈ H. Then φ(a) = a and φ(b) = b by definition of H. Then

φ(ab) = φ(a)φ(b) = ab so ab ∈ H. If a ∈ H so that φ(a) = a, then φ(a−1) =(φ(a)

)−1= a−1 so

a−1 ∈ H. Thus by the 2 Step Subgroup Test we see that H is a subgroup of G.

26. Prove that the mapping from U(16) to itself given by x 7→ x3 is an automorphism. What aboutx 7→ x5 and x 7→ x7? Generalize.

Recall that U(16) = {1, 3, 5, 7, 9, 11, 13, 15}. We will make a table of values for the functionf(x) = x3 in U(16).

x x3

1 13 115 137 79 911 313 515 15

It is clear from this table that f(x) = x3 is a bijection, that is, both 1-1 and onto. We canshow OP by recalling how modular arithmetic works: f(xy) ≡ (xy)3 ≡ x3y3 ≡ f(x)f(y). Thus

f(x) = x3 is an automorphism of U(16).

For x5 and x7 we can also make tables:

x x5 x x7

1 1 1 13 3 3 115 5 5 137 7 7 79 9 9 911 11 11 313 13 13 515 15 15 15

It is clear that both of these are also automorphisms of U(16) since x5 = x is the identity func-tion, and x7 = x3 which we just showed is an automorphism.

Generalize: It seems that any odd power of x, x2k+1, will always be an automorphism of U(16).We can prove this by noting the table of values for x4:

x x4

1 13 15 17 19 111 113 115 1

Since x4 = 1 in U(16), we see that x4k+1 = (x4)kx1 = x and x4k+3 = (x4)kx3 = x3 which wealready mentioned are automorphisms. Since any odd number is of the form 4k + 1 or 4k + 3,we see that x2k+1 for any k ∈ N is an automorphism of U(16) (note however that we have onlydiscussed two distinct functions, even if the formula is different).

You can also note that any even power of x will not be an automorphism since (1)2k = 1 and(15)2k = (−1)2k = 1 in U(16).

35. Show that the mapping φ(a+ bi) = a− bi is an automorphism of the group of complex numbersunder addition (C). Show that φ preserves multiplication as well – that is, φ(xy) = φ(x)φ(y) forall x and y in C.

Let a + bi, c + di ∈ C. If φ(a + bi) = φ(c + di) then a − bi = c − di and by the properties ofcomplex numbers that means that a = c and −b = −d which means b = d. Thus a+ bi = c+ diand φ is 1-1.

Let a+ bi ∈ C. Then a− bi ∈ C and φ(a− bi) = a− (−b)i = a+ bi so φ is onto.

Let a+ bi, c+ di ∈ C. Then φ((a+ bi) + (c+ di)

)= φ

((a+ c) + (b+ d)i

)= (a+ c)− (b+ d)i =

a+ c− bi− di = (a− bi) + (c− di) = φ(a+ bi) + φ(c+ di). So φ is operation preserving (OP).

Therefore φ is an automorphism of C.

OP for multiplication: Let a+ bi, c+ di ∈ C. Then φ((a+ bi)(c+ di)

)= φ(ac+ adi+ bci− bd) =

φ((ac− bd) + (ad+ bc)i

)= (ac− bd)− (ad+ bc)i. At this point it will be easier to show that we

get the same value if we consider φ(a + bi)φ(c + di) = (a − bi)(c − di) = ac − adi − bci − bd =(ac− bd)− (ad+ bc)i. Thus φ preserves multiplication as well.

55. Suppose that φ is an automorphism of D4 such that φ(R90) = R270 and φ(V ) = V . Determineφ(D) and φ(H).

Since R90V = D we have φ(D) = φ(R90V ) = φ(R90)φ(V ) = R270V = D′.

Since R90D = H we have φ(H) = φ(R90D) = φ(R90)φ(D) = R270D′ = H.

56. In Aut(Z9), let αi denote the automorphism that sends 1 to i where gcd(i, 9) = 1. Write α5 andα8 as permutations of {0, 1, . . . , 8} in disjoint cycle form.

Recall that since Z9 = 〈1〉, we have αi(k) = kαi(1) = ki. We will make a table of values for eachfunction, then convert to disjoint cycle form.

x α5(x) = 5x0 01 52 13 64 25 76 37 88 4

So α5 = (1 5 7 8 4 2)(3 6).

x α8(x) = 8x = −x = 9− x0 01 82 73 64 55 46 37 28 1

So α8 = (1 8)(2 7)(3 6)(4 5).