Just seeing if i'm right:

With the set of solutions for $z^4=1$: $\xi_4=\{1,i,-1,-i\}$, one can construct the group of the $4$th roots of unity: $(\xi_4,\cdot_\mathbb{C})$ and its multiplicative subgroup $(\xi_4^*,\cdot_\mathbb{C})$ with $\xi_4^*=\{1,-1\}$.

What is the Index $[ \xi_4 : \xi_4^* ]$?

I'd say through Lagrange: $$ [ \xi_4 : \xi_4^* ] =\frac{|\xi_4|}{|\xi_4^*|}=\frac{4}{2}=2.$$ Or counting the cosets of $\xi_4^*$: $$1\cdot\xi_4^*=\{1,-1\}= -1\cdot\xi_4^*,\quad i\cdot\xi_4^*=\{i,-i\}= -i\cdot\xi_4^*$$ we get two.

Is this correct?

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    $\begingroup$ What does $\xi_4$ mean? $\endgroup$ – Zev Chonoles May 15 '13 at 23:56
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    $\begingroup$ So $\xi_4$ is $4$th roots of unity, and $\xi_4^*$ is the square roots of unity? $\endgroup$ – Sammy Black May 15 '13 at 23:58
  • $\begingroup$ sorry, I missed to clarify that... $\endgroup$ – scjorge May 16 '13 at 0:00

Yes, indeed, the index $$[\xi_4:\xi^*_4] = \frac{|\xi_4|}{|\xi^*_4|} = \dfrac 4 2 = 2$$

Alternatively, as you've shown, the index $=$ the number of cosets of $\;\xi^*_4 \leq \xi_4 = 2$.

When you're dealing with a finite group, as in this case, the first "method" of computing the index of a subgroup with respect to its containing group is probably quickest. (Indeed, the first method works strictly for finite groups!)


Yes, it looks good. There are $2$ cosets of the subgroup $\xi_4^*$ in $\xi_4$.


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