The group of positive rational numbers under multiplication is a well-known mathematical structure. In this response, we will explore whether this group is isomorphic to a proper subgroup of itself.



To determine if the group of positive rational numbers under multiplication is isomorphic to a proper subgroup of itself, we first need to understand what isomorphism means in the context of group theory.

An isomorphism is a bijective map between two groups that preserves the group structure. In other words, if two groups A and B are isomorphic, there exists a function f: A → B that is one-to-one, onto, and satisfies the following condition:

f(ab) = f(a)f(b) for all a, b ∈ A

If we can find such a function between the group of positive rational numbers under multiplication and a proper subgroup of itself, then the two groups are isomorphic.



The group of positive rational numbers under multiplication, denoted by Q^+, consists of all positive fractions of the form p/q, where p and q are integers with q ≠ 0. Multiplication is the group operation.

To determine if Q^+ is isomorphic to a proper subgroup of itself, we need to find a proper subgroup of Q^+ and a function between them that satisfies the isomorphism condition.



A proper subgroup of a group is a subgroup that is not equal to the whole group. In the case of Q^+, we are looking for a subgroup H such that H ≠ Q^+.

One example of a proper subgroup of Q^+ is the subgroup generated by a single positive rational number. Let's consider the subgroup generated by 2/3.

The subgroup generated by 2/3 is {2/3, 4/9, 8/27, ...}. It consists of all positive rational numbers that can be obtained by multiplying powers of 2/3.



Now, let's define a function f: Q^+ → H that maps each positive rational number in Q^+ to its corresponding element in the subgroup generated by 2/3.

For example, f(2/5) = (2/3)^(-1) * (2/5) = 10/9.

We can verify that this function satisfies the isomorphism condition:

f(ab) = f(a)f(b) for all a, b ∈ Q^+

Let's consider two elements a = p/q and b = r/s, where a, b ∈ Q^+.

f(a)f(b) = (2/3)^(p/q) * (2/3)^(r/s) = (2/3)^(p/q + r/s) = (2/3)^(ps + qr)/(3^rs)

On the other hand, f(ab) = (2/3)^(p/q * r/s) = (2/3)^(pr/qs) = (2/3)^(pr/qs)/(3^rs)

Therefore, f(ab) = f(a)f(b), and the function f satisfies the isomorphism condition.



In conclusion, we have found a proper subgroup of the group