MAT 3702 Assignment 1 2026
Due Date: 13 May 2026
1. Let A, B, C be sets and show that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
Question 1
(⊆) Let x ∈ A ∪ (B ∩ C).
Then x ∈ A or x ∈ B ∩ C .
If x ∈ A, then x ∈ A ∪ B and x ∈ A ∪ C . Hence x ∈ (A ∪ B) ∩ (A ∪ C).
If x
∈ B ∩ C , then x ∈ B and x ∈ C . Hence x ∈ A ∪ B and x ∈ A ∪ C , so x ∈ (A ∪ B) ∩
(A ∪ C).
In both cases, x ∈ (A ∪ B) ∩ (A ∪ C). Thus A ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (A ∪ C).
(⊇) Let x ∈ (A ∪ B) ∩ (A ∪ C).
Then x ∈ A ∪ B and x ∈ A ∪ C .
If x ∈ A, then clearly x ∈ A ∪ (B ∩ C).
If x ∈/ A, then because x ∈ A ∪ B we must have x ∈ B . Similarly, from x ∈ A ∪ C we must
have x ∈ C . Therefore x ∈ B ∩ C , and hence x ∈ A ∪ (B ∩ C).
In both cases, x ∈ A ∪ (B ∩ C). Thus (A ∪ B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C).
each set contains the other, they are equal:
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
, Question 2
Let G be a finite group with x, y ∈ G. Prove that xy and yx have the same order.
We show that for any integer k ≥ 1, (xy)k = e if and only if (yx)k = e. This will imply that the
orders are equal.
First note the conjugation relation: yx = y(xy)y −1 . Indeed,
y(xy)y −1 = (yx)(yy −1 ) = yx.
Therefore, for any positive integer k ,
(yx)k = (y(xy)y −1 )k = y(xy)k y −1 .
Now if (xy)k = e, then
(yx)k = yey −1 = e.
if (yx)k = e, then
e = (yx)k = y(xy)k y −1 ⟹ y(xy)k = y ⟹ (xy)k = e.
Thus (xy)k = e ⟺ (yx)k = e. The smallest positive k for which either power becomes the identity
is identical; hence the orders of xy and yx are equal.
Due Date: 13 May 2026
1. Let A, B, C be sets and show that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
Question 1
(⊆) Let x ∈ A ∪ (B ∩ C).
Then x ∈ A or x ∈ B ∩ C .
If x ∈ A, then x ∈ A ∪ B and x ∈ A ∪ C . Hence x ∈ (A ∪ B) ∩ (A ∪ C).
If x
∈ B ∩ C , then x ∈ B and x ∈ C . Hence x ∈ A ∪ B and x ∈ A ∪ C , so x ∈ (A ∪ B) ∩
(A ∪ C).
In both cases, x ∈ (A ∪ B) ∩ (A ∪ C). Thus A ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (A ∪ C).
(⊇) Let x ∈ (A ∪ B) ∩ (A ∪ C).
Then x ∈ A ∪ B and x ∈ A ∪ C .
If x ∈ A, then clearly x ∈ A ∪ (B ∩ C).
If x ∈/ A, then because x ∈ A ∪ B we must have x ∈ B . Similarly, from x ∈ A ∪ C we must
have x ∈ C . Therefore x ∈ B ∩ C , and hence x ∈ A ∪ (B ∩ C).
In both cases, x ∈ A ∪ (B ∩ C). Thus (A ∪ B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C).
each set contains the other, they are equal:
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
, Question 2
Let G be a finite group with x, y ∈ G. Prove that xy and yx have the same order.
We show that for any integer k ≥ 1, (xy)k = e if and only if (yx)k = e. This will imply that the
orders are equal.
First note the conjugation relation: yx = y(xy)y −1 . Indeed,
y(xy)y −1 = (yx)(yy −1 ) = yx.
Therefore, for any positive integer k ,
(yx)k = (y(xy)y −1 )k = y(xy)k y −1 .
Now if (xy)k = e, then
(yx)k = yey −1 = e.
if (yx)k = e, then
e = (yx)k = y(xy)k y −1 ⟹ y(xy)k = y ⟹ (xy)k = e.
Thus (xy)k = e ⟺ (yx)k = e. The smallest positive k for which either power becomes the identity
is identical; hence the orders of xy and yx are equal.