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The hypotheses make $K$ a CM-field, with nontrivial automorphism $c$ over the totally real number field $k$. In every complex embedding of $K$, this automorphism acts as complex conjugation. Write $E_K={\mathcal{O}}_K^{\times }$, $E_k={\mathcal{O}}_k^{\times }$ and ${\mu }_K$ for the finite cyclic group of roots of unity in $K$.

For any $u\in E_K$, the element $u/c(u)$ is an algebraic integer, indeed a unit, and every conjugate has absolute value one. The Kronecker theorem on algebraic integers in the unit disk makes it a root of unity. For clarity, the essential proof is finite: the conjugates of all positive powers have modulus one, so the integer coefficients of their monic minimal polynomials are bounded by fixed binomial coefficients. There are only finitely many such polynomials and roots; two powers coincide, forcing a root of unity.

Thusis a homomorphism. Its kernel is exactly $E_k$, since a conjugation-fixed element lies in $k$ and both it and its inverse are integral there. For $\zeta \in {\mu }_K$, $c(\zeta )={\zeta }^{-1}$ and $\varphi (\zeta )={\zeta }^2$. Consequently $\varphi$ induces an injectionTo check injectivity explicitly, if $\varphi (u)={\zeta }^2$, then $\varphi (u/\zeta )=1$, so $u/\zeta \in E_k$. The cyclic group ${\mu }_K$ has even order because it contains $-1$, and its quotient by squares has order two. Therefore the unit index of a CM-field satisfies

First suppose $|{\sigma }_i(\alpha )|=1$ for every complex embedding. The coefficients of the monic polynomialare rational algebraic integers and hence integers. They are uniformly bounded in $m$, because each is an elementary symmetric sum of numbers of modulus one. Only finitely many such integer polynomials can occur, so only finitely many algebraic integers ${\alpha }^m$ occur. Two powers coincide; since $\alpha \ne 0$, this gives ${\alpha }^{r-s}=1$. Thus $\alpha$ is a root of unity. This is Kronecker theorem on algebraic integers in the unit disk.

The second printed assertion is literally false for $\alpha =0$. For the intended statement with $\alpha \ne 0$, use the Minkowski embedding: algebraic integers of $K$ form a lattice, so only finitely many have all conjugates of modulus at most $2$. A nonzero algebraic integer with every conjugate of modulus at most one hasso every conjugate has modulus one and the first part applies. Consequently every non-root in that finite box has ${\max }_i|{\sigma }_i(\alpha )|>1$. Choose $c>1$ no larger than the least of these finitely many maxima, taking $c=2$ if there are no non-roots. Then every nonzero $\alpha \in {\mathcal{O}}_K$ satisfying $|{\sigma }_i(\alpha )|<c$ for all $i$ is a root of unity.

Dirichlet unit theorem states that for signature $(r_1,r_2)$,and the logarithmic embedding of number field units maps the free part to a full lattice in the hyperplane where the weighted coordinate sum is zero.

For a real quadratic field, the roots of unity are $\{\pm 1\}$ and the logarithms of positive units form a nonzero discrete subgroup $a\mathbb{Z}\subset \mathbb{R}$. Its smallest positive element is $\log ϵ$ for a unit $ϵ>1$. Given any positive unit $u>1$, choose $n$ so that $1\le u{ϵ}^{-n}<ϵ$; minimality forces the quotient to be $1$. Applying signs and inverses shows

Since $11\equiv 3(\mathrm{mod}4)$, the ring of integers of a quadratic field is $\mathbb{Z}[\sqrt{11}]$. A unit $a+b\sqrt{11}$ satisfies the Pell equation $a^2-11b^2=\pm 1$. The elementhas norm $1$. To prove it is the smallest unit above one without invoking the continued-fraction algorithm, suppose $1<a+b\sqrt{11}<10+3\sqrt{11}$. Replacing by its inverse or negative if needed makes $b>0$, and the conjugate relation implies $b\sqrt{11}<(u+u^{-1})/2<10$, so $b\le 3$. For $b=1,2,3$, the numbers $11b^2\pm 1$ are respectivelyOnly the last list contains a square, namely $100=10^2$, which yields the endpoint itself. Hence no smaller unit exists and

A nonabelian simple group has no nonidentity conjugacy class of prime-power size. Coprime-degree irreducible characters vanish on such an element: the conjugacy-class sum makes the character-to-degree ratio an algebraic integer, and the Kronecker theorem on algebraic integers in the unit disk makes a nonzero ratio a root of unity, forcing a scalar in a faithful group representation. Column character orthogonality then makes $1/p$ an algebraic integer, impossible. This is the character-theoretic ingredient in Burnside's theorem.

For a CM-field, put $c$ for conjugation and $E_K={\mathcal{O}}_K^{\times }$. The map $u\mapsto u/c(u)$ takes values in the cyclic group ${\mu }_K$ of roots of unity: all conjugates have absolute value one, and the Kronecker theorem on algebraic integers in the unit disk applies. Its kernel is $E_{K^{+}}$, while its image on ${\mu }_K$ is ${\mu }_K^2$. It therefore embeds $E_K/(E_{K^{+}}{\mu }_K)$ into ${\mu }_K/{\mu }_K^2$, a group of order two. This proves the index bound without assuming that every unit is a real unit times a root of unity.