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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.

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    $\begingroup$ 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. $\endgroup$ – Ben Sheller Oct 19 '17 at 0:18
  • $\begingroup$ Hint to get started: Any element $z$ of a subgroup of order $4$ must satisfy $z^4 = 1$, hence $|z|^4 = 1$, hence $|z| = 1$. $\endgroup$ – Bungo Oct 19 '17 at 0:26
  • $\begingroup$ Ok, I posted some of my work. $\endgroup$ – Cody S Oct 19 '17 at 0:28
  • $\begingroup$ 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. $\endgroup$ – Matt Samuel Oct 19 '17 at 3:25
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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, $-1$).

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