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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 $(\mathbb{Q}^*)^2 = \{ q^2 | q \in \mathbb{Q^*} \}$.

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  • $\begingroup$ Yes. I will fix that $\endgroup$ – user486995 Sep 24 '20 at 16:36
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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 $q_1,\dots,q_k$.

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  • $\begingroup$ I am not familiar with the notation $\mathbb{Z}_2^{+ \mathbb{N}}$. $\endgroup$ – user486995 Sep 24 '20 at 16:44
  • $\begingroup$ It is the direct sum $\bigoplus_{i\in\mathbb N}\mathbb Z_2$ (see en.wikipedia.org/wiki/Direct_sum_of_groups) $\endgroup$ – Kenta S Sep 24 '20 at 16:44
  • $\begingroup$ gotcha. What would the generators be of the original quotient group? $\endgroup$ – user486995 Sep 24 '20 at 16:45
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    $\begingroup$ $-1$ and the prime numbers. $\endgroup$ – Kenta S Sep 24 '20 at 16:46
  • $\begingroup$ I guess i'm still a little confused. What would an element of the quotient group look like. $\endgroup$ – user486995 Sep 24 '20 at 21:15

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