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Math 120, Solution Set 3
1.7:1. If x, y ∈ F × and a ∈ F , then
x · (y · a) = x · (ya) = x(ya) = (xy)a = (xy) · a
by the definition of the action of F × on F and the fact that multiplication in a field
is associative. Similarly, 1 · a = a since 1 is the identity element for multiplication
in F . Thus the action of F × on F is a group action.
1.7:3. Suppose r, s ∈ R and (x, y) ∈ R2 . Then
r · (s · (x, y)) = r · (x + sy, y) = ((x + sy) + ry, y)
= (x + (r + s)y, y) = (r + s) · (x, y),
0 · (x, y) = (x + 0y, y) = (x, y)
(note that we are using the additive group R, so the identity element is 0) so this
is a group action.
1.7:12. Let A be the set of pairs of opposite vertices. If {a, b} ∈ A and g ∈ D2n , let
g · {a, b} = {g(a), g(b)}.
(*)
By elementary geometry, g(a) and g(b) are also opposite. Thus (*) does define a
map from D2n × A to A. If 1 is the identity element in D2n , i.e., the identity map
on vertices, then 1 · {a, b} = {1(a), 1(b)} = {a, b}. Also, if g and h are two elements
of D2n , then
g ·(h·{a, b}) = g ·{h(a), h(b)} = {g(h(a)), g(h(b))} = {(gh)(a), (gh)b} = (gh)·{a, b}.
Thus · is a group action.
To find the kernel K of the action, let’s label the vertices 1, . . . , n. Since n is even,
n = 2k for some k.
Recall that every element of D2n can be expressed as ri sj for 0 ≤ i < n and 0 ≤
j ≤ 1. Suppose ri sj is in the kernel of the action. Consider the pair p = {1, k + 1}
of opposite vertices. Now ri sj maps vertex 1 to vertex 1 + i, so 1 + i must belong
to the pair p. That is, 1 + i must be 1 or k + 1, so i must be 0 or k.
Thus the only elements that might belong to K are r0 , rk , r0 s, and rk s. That is,
1, rk , s, and rk s. Now 1 is in the kernel, as is rk (since rk maps each vertex to the
opposite vertex.)
Now consider the action of s on the pair {2, k + 2}:
s · {2, k + 2} = {s(2), s(k + 2)} = {n, k}.
1
2
So s is not in the kernel unless 2 is n or k, i.e., unless n = 2 or n = 4. In these
cases, there are no other pairs, so s is indeed in K. Since rk and s are in K, so is
rk s.
However, for n > 4, s ∈
/ K. Therefore rk s cannot be in K. Thus for n > 4, the
kernel contains just 1 and rk .
1.7:18. Suppose a, b, c ∈ A. Then 1 ∈ H and a = 1 · a by definition of group action,
so a ∼ a and ∼ is reflexive.
If a ∼ b then a = hb for some h ∈ H, so h−1 ∈ H and b = h−1 a, and so b ∼ a and
∼ is symmetric.
Finally, suppose a ∼ b and b ∼ c. Then a = h1 b and b = h2 c for some h1 , h2 ∈ H,
so (h1 h2 )c = h1 (h2 c) = h1 b = a by definition of group action. Therefore a ∼ c so
∼ is transitive, and thus ∼ is an equivalence relation.
2.1:5. Suppose H ⊂ G and |H| = n − 1, say H = G − {g} for some g ∈ G.
Choose a ∈ H, a 6= 1 (this is possible since |G| > 2, so |H| > 1). Since a 6= 1,
a−1 6= 1, and so (by the cancellation property) a−1 g 6= g. Therefore a−1 g ∈ H, but
a(a−1 g) = g ∈
/ H, so H cannot be a subgroup.
2.1:6. Let H be the subset consisting of all elements of finite order. Of course
1 ∈ H, so H is not empty. Also, if x ∈ H, then x−1 ∈ H since |x| = |x−1 |. Finally,
if x, y ∈ H then xn = y m = 1 for some n, m ∈ Z+ . But then (since G is abelian)
(xy)( nm) = xnm y nm = (xn )m (y m )n = 1
so xy ∈ H. This proves that H is a subgroup.
Example: consider the group of translations and rotations of the plane. Rotations
by 180 degrees have order 2, but the product of two such rotations (with different
centers) is a translation of infinite order.
2.2:2. If z ∈ Z(G) and g ∈ G, then zg −1 z −1 = g −1 by definition of Z(G). Multiplying on the left by g and on the right by z we conclude that gzg −1 = z,
so g ∈ CG (Z(G)). Therefore CG (Z(G)) = G, and CG (Z(G)) ≤ NG (Z(G)) so
NG (Z(G)) = G as well.
2.2:10. Suppose H = {1, h} and g ∈ NG (H). Then gHg −1 = H, so ghg −1 ∈ H.
But ghg −1 6= 1 (or else h = 1), so ghg −1 = h. Trivially g1g −1 = 1, so g ∈ CG (H).
We always have CG (H) ≤ NG (H), so CG (H) = NG (H).
If NG (H) = G then CG (H) = G, so g −1 hg = h for every g. Multiplying on the left
by g and on the right by h−1 gives hgh−1 = g, so h ∈ Z(G) and H ≤ Z(G).
2.3:10. By Proposition 5(2) (with x = 1), the order of 30 is 54/(54, 30) = 54/6 = 9.
The list of elements and their orders in h30i is (now using Proposition 5(2) with
3
x = 30, so the order of k · 30 is 9/(9, k))
k:
0
k · 30 0
order: 1
1 2
30 6
9 9
3
36
3
4
12
9
5
42
9
6
18
3
7
48
9
8
24
9
2.4:14. (a). G =< G >.
(b). Z =< 1 >.
(c). Let H and k be as in the hint. Then each generator has the form a/b where b
is a factor of k, so k = bc for some c. But then
ac
ac
1
a
=
=
= ac
b
bc
k
k
so a/b is the cyclic group C generated by 1/k. Since each generator of H is in C,
in fact H is contained in C. Since each subgroup of a cyclic group is cyclic, in fact
H must be cyclic.
(d). Let H be a finitely generated subgroup of Z. By (c), it’s contained in the
cyclic group generated by 1/k. But 1/(2k) is not in this cyclic group and thus is
not in H. Thus H is not all of Q.
15. For instance, the subgroup of dyadic numbers, i.e. numbers of the form a/(2n ).
It’s a proper subgroup because 1/3 is not in it.