Problem session solutions
1. Let P = R2 = {
x
y
: x, y ∈ R} denote the plane, and consider the relation ∼ on P defined by
p ∼ q ⇐⇒ p and q are linearly dependent.
(a) Show that ∼ is not an equivalence relation on P .
(b) Show that ∼ is an equivalence relation on P \ { 00 }. Then compute the equivalence
classes and find a complete set of equivalence class representatives.
Solution (a) This isn’t transitive, since e1 ∼ 0 and 0 ∼ e2 , but e1 6∼ e2 .
(b) This is clearly reflexive and symmetric, and if p ∼ q and q ∼ r where p, q, r are all non-zero,
then p = tq and q = sr for some scalars t, s, so p = (st)r, so p ∼ r. The equivalence class
of p is [p] = {tp : t ∈ R, t 6= 0} and a complete set of equivalence class representatives is
θ
{ cos
sin θ : 0 ≤ θ < π}. [Draw
a picture to justify this]. Another complete set of equivalence
1
class representatives is { t : t ∈ R} [draw another picture]. And there are lots more.
2. Let H be a permutation group on a non-empty set S, and consider the relation ∼ of H-orbit
equivalence on S defined by
x ∼ y ⇐⇒ ∃ α ∈ H : α(x) = y.
Theorem 20 shows that ∼ is an equivalence relation on S. Compute the equivalence classes
and find a complete set of equivalence class representatives for ∼ if:
(a) S = {1, 2, 3, 4}, H = S4
(b) S = {1, 2, 3, 4}, H = {(1)}
(c) S = {1, 2, 3, 4, 5}, H = {(1), (2 3), (4 5)}
(d) S = P = R2 , H = {τa : a = 2n
0 , n ∈ Z}
(e) S = P = R2 , H = {τa , τb ◦ r : a = 2n
0 ,b =
2m+1
0
, n, m ∈ Z}
Solution (a) Equivalence classes: {S}; CSECR’s: {1}.
(b) Equivalence classes: {{1}, {2}, {3}, {4}}; CSECR’s: S.
(c) Equivalence classes: {{1}, {2, 3}, {4, 5}}; CSECR’s: {1, 2, 4}.
x
(d) Equivalence classes: {[p] = {p + 2n
0 : n ∈ Z} : p ∈ P }; CSECR’s: { y : 0 ≤ x < 2, y ∈
R}.
2n+1
(e) Equivalence classes: {[p] = {p+ 2n
: n ∈ Z} : p ∈ P }; CSECR’s: { xy : 0 ≤
0 , r(p)+
0
x < 2, y ≥ 0}.