What does $k^*/k^{*^2}$ mean?

Question

I'm trying to get a more concrete understanding of what these elements 'look like.'

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Here $k$ is a field, $k^*$ is multiplicative group, and $(k^*)^2$ consists of the squares in $k^*$.

Answer 1

Presumably, $k$ is a field. In this case, $k^*$ is its multiplicative groups and $k^{*^2}$ is the subgroup of squares.

What the quotient group $k^*/k^{*^2}$ looks like depends very much on $k$.

For $k=\mathbb C$, $\quad k^*/k^{*^2}$ is the trivial group.

For $k=\mathbb R$, $\quad k^*/k^{*^2}$ is the cyclic group of order $2$.

For $k=\mathbb Q$, $\quad k^*/k^{*^2}$ can be described by the square-free fractions, and so is the set of all finite products of primes to the power $\pm1$, but I don't know which concrete group it is.