Let $U_{12} = \{a + ib: a,b\in \mathbb{R} \,\, \wedge \,\, (a + ib)^{12} = 1\}$. Find all the subgroups of order 2 and also the subgroups of index 2.

The only subgroup of order 2 that I can think of would be $\{-1,1\}$ since we need an identity element, and an element that is it's own inverse.

As for the subgroups of index 2, I think it would just be $U_6$ defined similarly as $U_{12}$. Are these it, or am I missing something?

  • 1
    $\begingroup$ You are missing nothing. $\endgroup$
    – ajotatxe
    Oct 25, 2016 at 13:35
  • $\begingroup$ Indeed, $U_{12}$ is isomorphic to the cyclic group of order 12, hence you are right about the subgroups being what you claim. $\endgroup$ Oct 25, 2016 at 14:19

1 Answer 1


You can rewrite the definition as $U_{12} = \{ x \in \mathbb{C} ,\; x^{12}=1\}$. The fundamental theorem of algebra tells you that that equation has exactly $12$ solutions in $\mathbb{C}$.

Using the standard way of solving this kind of equation, you get that $$e^{k\frac{i\pi}{6}}$$ is a solution for $k=0, \dots, 11$ (actually for every $k$ integer, but then they start repeating).

All this proves that $U_{12}$ is isomorphic to the cyclic group of order $12$, since it is generated by $e^{i\frac{\pi}{6}}$.

Now, a subgroup of index $2$ has to have $6$ elements. Easily, it is generated by $\sigma^2$. The subgroup of order $2$ is generated by $\sigma^6$ (which happens to be equal to $-1$ in $\mathbb{C}$, as you correctly said).

You can easily verify by direct inspection that any other element has order $4$ or $12$, and that there are no other subgroups.


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