Multiplicative Group of Rationals is Subgroup of Complex

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Theorem

Let $\left({\Q, \times}\right)$ be the multiplicative group of rational numbers.

Let $\left({\C, \times}\right)$ be the multiplicative group of complex numbers.


Then $\left({\Q, \times}\right)$ is a normal subgroup of $\left({\C, \times}\right)$.


Proof

From Multiplicative Group of Rationals is Subgroup of Reals, $\left({\Q, \times}\right) \lhd \left({\R, \times}\right)$.

From Multiplicative Group of Reals is Subgroup of Complex, $\left({\R, \times}\right) \lhd \left({\C, \times}\right)$.

Thus $\left({\Q, \times}\right) \le \left({\C, \times}\right)$.

From Non-Zero Complex Numbers under Multiplication form Abelian Group, $\left({\C, \times}\right)$ is abelian.

From Subgroup of Abelian Group is Normal it follows that $\left({\Q, \times}\right) \lhd \left({\C, \times}\right)$.

$\blacksquare$