Proof:

Given:
Let G be a group and x, y ∈ G.
Given that the order of x is 5, i.e., |x| = 5.
Also, given that xy = y^(-1)x, where y is not equal to the identity element.

To Find:
We need to find the order of y, i.e., |y|.

Proof:
Step 1: Proof that y is not equal to the identity element.
Assume that y is equal to the identity element.
Then xy = y^(-1)x becomes x = y^(-1)x, which implies y^(-1) = e (identity element).
Multiplying both sides by y, we get y^(-1)y = ey, which simplifies to e = ey.
This means that e = y, which contradicts our assumption that y is not equal to the identity element.
Hence, y cannot be equal to the identity element.

Step 2: Proof that the order of y is a divisor of 5.
To prove this, let's consider the subgroup generated by y, denoted by ⟨y⟩.
Since the order of x is 5, the subgroup generated by x, denoted by ⟨x⟩, has 5 distinct elements: e, x, x^2, x^3, and x^4.
Now, let's consider the set S = {xy, (xy)^2, (xy)^3, (xy)^4}.
Using the given condition that xy = y^(-1)x, we can rewrite S as S = {y^(-1)x, (y^(-1)x)^2, (y^(-1)x)^3, (y^(-1)x)^4}.
Simplifying, we get S = {y^(-1)x, x^2y^(-1), xy^(-1)x^2y^(-1), x^3y^(-1)x^2y^(-1)}.
Notice that S is a subset of the subgroup generated by x, i.e., S ⊆ ⟨x⟩.
Also, S is a subset of the subgroup generated by y, i.e., S ⊆ ⟨y⟩.
Since S is a subset of both ⟨x⟩ and ⟨y⟩, it must also be a subset of their intersection, denoted by ⟨x⟩ ∩ ⟨y⟩.
The intersection of two subgroups is also a subgroup, so ⟨x⟩ ∩ ⟨y⟩ is a subgroup of G.
Now, since S is a subset of ⟨x⟩ ∩ ⟨y⟩, the cardinality of S, denoted by |S|, must be less than or equal to the cardinality of ⟨x⟩ ∩ ⟨y⟩, denoted by |⟨x⟩ ∩ ⟨y⟩|.
Since |S| ≤ |⟨x⟩ ∩ ⟨y⟩| and |S| = 4, we have 4 ≤ |⟨x⟩ ∩ ⟨y⟩|.

Step 3: Proof that the order of y is not equal to 5.
Assume that the order of y is 5, i.e., |y| = 5.
This implies that ⟨y⟩ is a cyclic subgroup of G with