# determining finite subgroups of a multiplicative group

This problem deals with finite subgroups of the multiplicative group of a field $(F^*)$.

Consider the field of complex numbers $\mathbb{C}$. How many subgroups of order 4 does $\mathbb{C}^*$ have?

Is this group cyclic?

Attempted work:

Consider the field of $\mathbb{C}$. Clearly, $\mathbb{C}^*$ has elements of order $1,2,3,4...$ with the subgroups of $\mathbb{C}^*$ also having orders $1,2,3,4...$ Now, Let $x \in \mathbb{C}^*$ with unity $x^4=1$. $$\Rightarrow x=(1)^{\frac{1}{4}}$$ $$\Rightarrow x^2=|x^2|=4$$ $$\Rightarrow x^3=|x^3|=4$$ $$\Rightarrow x^4=|x^4|=4$$

So the number of cyclic subgroups of order 4 in $\mathbb{C}^*$ is 4.

• What are your thoughts on the problem? What have you tried, and where are you stuck? Knowing this will make providing quality answers easier for other users. – Ben Sheller Oct 19 '17 at 0:18
• Hint to get started: Any element $z$ of a subgroup of order $4$ must satisfy $z^4 = 1$, hence $|z|^4 = 1$, hence $|z| = 1$. – Bungo Oct 19 '17 at 0:26
• Ok, I posted some of my work. – Cody S Oct 19 '17 at 0:28
• Do you know what a subgroup is? Do you know the specific complex numbers that are fourth roots of $1$? That's why they picked this example, you can compute it easily. – Matt Samuel Oct 19 '17 at 3:25

If $z$ is an element of a group $G \subseteq \Bbb C^{\ast}$ of order $4$, then $z$ is a root of $x^4 - 1$.
Since $\Bbb C$ is a field, there are at most four such roots. Since:
$x^4 - 1 = (x^2 + 1)(x^2 - 1) = (x + i)(x - i)(x + 1)(x - 1)$, we see there are precisely four such roots. These form a cyclic group of order $4$: $\langle i\rangle = \langle -i\rangle$.
Thus $\{1,i,-1,-i\}$ is the sole subgroup of order four in $\Bbb C^{\ast}$ (there is no subgroup isomorphic to $V$, since $\Bbb C^{\ast}$ has but one element of order $2$, namely, $-1$).