# Group structure of this quotient group

I'm curious about the quotient group $$\frac{\mathbb{Q}^* }{(\mathbb{Q}^*)^2}$$. What do its elements look like? Is it isomorphic to some infinite group?

** For clarification on notation, $$\mathbb{Q}^* = \mathbb{Q} \setminus \{0 \}$$, and $$(\mathbb{Q}^*)^2 = \{ q^2 | q \in \mathbb{Q^*} \}$$.

• Yes. I will fix that
– user486995
Sep 24, 2020 at 16:36

By unique factorization, $$(\mathbb Q^*,\times)\cong \mathbb Z^{\oplus\mathbb N}\times\mathbb Z_2$$. Thus, $$\mathbb Q^*/(\mathbb Q^*)^2\cong \mathbb Z_2^{\oplus\mathbb N}\times\mathbb Z_2\cong\mathbb Z_2^{\oplus\mathbb N}$$.

P.S.:

The isomorphism $$(\mathbb Q^*,\times)\cong \mathbb Z^{\oplus\mathbb N}\times\mathbb Z_2$$ is given by $$\mathbb Z^{\oplus\mathbb N}\times\mathbb Z_2\to\mathbb Q^*:((n_i)_{i\in\mathbb N},m)\mapsto (-1)^m\prod_{i\in\mathbb N}p_i^{n_i}$$, where $$p_i$$ is the $$i$$th prime number, which is well-defined since all but finitely many $$n_i$$s are $$0$$.

Each element $$x\in \mathbb Q^*/(\mathbb Q^*)^2$$ can thus be represented as $$\pm q_1q_2\cdots q_k$$ for some distinct prime numbers $$q_1,\dots,q_k$$.

• I am not familiar with the notation $\mathbb{Z}_2^{+ \mathbb{N}}$.
– user486995
Sep 24, 2020 at 16:44
• It is the direct sum $\bigoplus_{i\in\mathbb N}\mathbb Z_2$ (see en.wikipedia.org/wiki/Direct_sum_of_groups) Sep 24, 2020 at 16:44
• gotcha. What would the generators be of the original quotient group?
– user486995
Sep 24, 2020 at 16:45
• $-1$ and the prime numbers. Sep 24, 2020 at 16:46
• I guess i'm still a little confused. What would an element of the quotient group look like.
– user486995
Sep 24, 2020 at 21:15