Assignment 6c
2, 3, 11, 13, 14, 22, 26, 35, 55, 56
2. Find Aut(Z).
Since Z = h1i, by property 4 of Theorem 6.2 we know that any automorphism of Z must map
1 to another generator of Z. Also, since automorphisms preserve the operation, we see that for
any automorphism φ, φ(k) = kφ(1) (as in example 13 in the text). Since the only generators of
Z are 1, −1, that gives us two possibilities for automorphisms in Aut(Z), namely f (x) = x and
g(x) = −x.
The identity function f (x) = x is clearly an automorphism (and should have been proven to be
an 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 is
onto. 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 automorphisms, 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 mapping
φ(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
z 2 ∈ R+ and φ(z 2 ) = z 2 = 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 φg φh = φgh .
Recall that φa is the inner automorphism of G induced by a, that is: φa (x) = axa−1 . To show
that the given functions are equal, I will show
that for any x ∈ G they have the same output
value. Let x ∈ G. Then φg φh (x) = φg φh (x) = φg (hxh−1 ) = g(hxh−1 )g −1 = (gh)x(h−1 g −1 ) =
(gh)x(gh)−1 = φgh (x). Since x was an arbitrary element of G, we see that φg φh = 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 gives
different outputs. First we’ll compare φR0 to all the others. Note that φR0 (x) = R0 xR0−1 = x
since R0 is the identity. Since φR0 (H) = H and φR90 (H) = R90 HR270 = D0 R270 = V we see that
φR0 6= φR90 . Since φR0 (R90 ) = R90 and φH (R90 ) = HR90 H = DH = R270 we see that φR0 6= φH .
Also, since φD (R90 ) = DR90 D = V D = R270 we see that φR0 6= φD . Thus φR0 is distinct from
the other three inner automorphisms listed.
−1
Now compare φR90 to the remaining two automorphisms. Since φR90 (R90 ) = R90 R90 R90
= R90
and φH (R90 ) = HR90 H = DH = R270 and φD (R90 ) = DR90 D = V D = R270 we see that
φR90 6= φH and φR90 6= φD .
It remains to show that φH 6= φD . Since φH (H) = HHH −1 = H and φD (H) = DHD = R270 D =
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 these
functions 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, and
then we need to check if those functions give automorphisms. Recall from an earlier chapter that
the 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 an
automorphism.
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 5z
mod 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 see
that e ∈ H so H 6= ∅. Let a, b ∈ H. Then φ(a) = a and φ(b) = b by definition of H. Then
−1
φ(ab) = φ(a)φ(b) = ab so ab ∈ H. If a ∈ H so that φ(a) = a, then φ(a−1 ) = φ(a)
= 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 about
x 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 function
f (x) = x3 in U (16).
x x3
1 1
3 11
5 13
7 7
9 9
11 3
13 5
15 15
It is clear from this table that f (x) = x3 is a bijection, that is, both 1-1 and onto. We can
show OP by recalling how modular arithmetic works: f (xy) ≡ (xy)3 ≡ x3 y 3 ≡ f (x)f (y). Thus
f (x) = x3 is an automorphism of U (16).
For x5 and x7 we can also make tables:
x x7
x x5
1 1
1 1
3 11
3 3
5 5
5 13
7 7
7 7
9 9
9 9
11 3
11 11
13 13
13 5
15 15
15 15
It is clear that both of these are also automorphisms of U (16) since x5 = x is the identity function, 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 1
3 1
5 1
7 1
9 1
11 1
13 1
15 1
Since x4 = 1 in U (16), we see that x4k+1 = (x4 )k x1 = x and x4k+3 = (x4 )k x3 = x3 which we
already 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 only
discussed 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 numbers
under addition (C). Show that φ preserves multiplication as well – that is, φ(xy) = φ(x)φ(y) for
all 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 of
complex numbers that means that a = c and −b = −d which means b = d. Thus a + bi = c + di
and 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 . Determine
φ(D) and φ(H).
Since R90 V = D we have φ(D) = φ(R90 V ) = φ(R90 )φ(V ) = R270 V = D0 .
Since R90 D = H we have φ(H) = φ(R90 D) = φ(R90 )φ(D) = R270 D0 = H.
56. In Aut(Z9 ), let αi denote the automorphism that sends 1 to i where gcd(i, 9) = 1. Write α5 and
α8 as permutations of {0, 1, . . . , 8} in disjoint cycle form.
Recall that since Z9 = h1i, we have αi (k) = kαi (1) = ki. We will make a table of values for each
function, then convert to disjoint cycle form.
x
0
1
2
3
4
5
6
7
8
α5 (x) = 5x
0
5
1
6
2
7
3
8
4
So α5 = (1 5 7 8 4 2)(3 6).
x α8 (x) = 8x = −x = 9 − x
0
0
8
1
2
7
3
6
4
5
5
4
6
3
7
2
8
1
So α8 = (1 8)(2 7)(3 6)(4 5).