Let $F$ be an infinite field such that $F^*$ is a torsion group. We know that $F^*$ is an Abelian group. So every subgroup of $F^*$ is a normal subgroup.

My question:

Does $F^*$ have a proper subgroup with finite index?

  • $\begingroup$ What is $F^*$? It can mean many things. $\endgroup$ – Patrick Da Silva Jun 21 '13 at 6:30
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    $\begingroup$ @PatrickDaSilva $F^*$ is a multiplicative group of field $F$. $\endgroup$ – Babak Miraftab Jun 21 '13 at 6:36
  • $\begingroup$ You mean the group $F^{\times} = F \backslash \{0\}$ with multiplication? I've never seen the notation $F^*$ before so I asked. Thanks $\endgroup$ – Patrick Da Silva Jun 21 '13 at 7:43
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    $\begingroup$ @PatrickDaSilva, I think that's a rather pretty standard notation in most books I remember. $\endgroup$ – DonAntonio Jun 21 '13 at 8:05
  • $\begingroup$ @Don : I guess we just didn't use the same books then! $\endgroup$ – Patrick Da Silva Jun 22 '13 at 4:08

Let $F$ be the algebraic closure of a finite field. Each element of $F^*$ belongs to a finite field, so is a torsion element. On the other hand $F^*$ cannot have a subgroup of index $n>1$. For if $A$ is such a subgroup, then $x^n\in A$ for all $x\in F^*$. But if $z\in F^*\setminus A$, then $z$ has an $n$th root in $F^*$ contradicting the previous sentence.

Returning to the general case. If $F^*$ is torsion, then obviously $F$ has finite characteristic, and is algebraic over its prime field. Therefore $F$ is contained in an algebraic closure of a finite field. If $F$ itself is finite, then it obviously has finite index subgroups, but this was excluded by the OP.

And as an example of an infinite field such that $F^*$ has a subgroup of a finite index let's try the following. Consider the nested union of extensions of $\mathbb{F}_2$ of degrees $2^n$ $$ F=\bigcup_{n\ge0}\mathbb{F}_{2^{2^n}}. $$ The union can be formed inside an algebraic closure of $\mathbb{F}_2$. For $m\ge n$ let $N^m_n:\mathbb{F}_{2^{2^m}}\to\mathbb{F}_{2^{2^n}}$ be the relative norm map. For $n\ge1$ define the groups $$ A_n=\{z\in\mathbb{F}_{2^{2^n}}\mid N^n_1(z)=1\}\le \mathbb{F}_{2^{2^n}}^*. $$ Transitivity of norm in a tower of field extensions means that $N^n_1\circ N^m_n=N^m_1$ for all $1\le n\le m$.

Let us define $$ A=\bigcup_{n\ge1}A_n. $$ I claim that $A$ is a subgroup of $F^*$. It is obviously closed under inverses, as all the $A_n$ are groups as kernels of $N^n_1$. If $z\in A_n$ and $n<m$, then $$ N^m_1(z)=N^n_1(N^m_n(z)). $$ Here $N^m_n(z)=z^{2^{m-n}}$ is just a power of $z$, so we get that also $z\in A_m$. We have seen that $A_n\le A_m$, and the claim follows from this.

To close off this example I claim that $A$ is of index three in $F^*$. Let $\mathbb{F}_4^*=\{1,\omega,\omega^2=1+\omega\}$, where $\omega$ is a primitive third root of unity. Because $N^n_1(\omega)=\omega^{2^{n-1}}$ is either $\omega$ or $\omega^2$, it follows that for every element $z\in F^*$ exactly one of $z,\omega z,\omega^2 z$ belongs to the subgroup $A$. The claim follows from this.

Note that the argument from the case of an algebraically closed field does not apply for this $F$. For example, the field $F$ does not have ninth roots of unity because those reside in the field $\mathbb{F}_{64}$, and won't be included in this tower.

  • $\begingroup$ What do you mean by "If F itself is finite, then it obviously has finite index subgroups" $\endgroup$ – Babak Miraftab Jun 21 '13 at 6:47
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    $\begingroup$ Well, the index of the trivial subgroup in the whole group is finite, say...:) $\endgroup$ – DonAntonio Jun 21 '13 at 8:06
  • $\begingroup$ @Babgen: That was an exaggeration in the sense that there may not be non-trivial subgroups. If $2^m-1$ is a (Mersenne) prime, then $F^*$ has no non-trivial subgroups. Other than those and the field of three elements the multiplicative group of a finite field always has non-trivial subgroups. $\endgroup$ – Jyrki Lahtonen Jun 21 '13 at 8:28

Note: This uses the very same reasoning as Jyrki's answer but with a different, perhaps slightly more groupwise, approach:

(1) An abelian group $\,A\,$ (with multiplicative operation, to fit within our problem) is divisible if

$$\forall\,g\in G\;\wedge\;\forall n\in\Bbb N\;\exists\, x\in G\;\;s.t.\;\;g=x^n$$

(2) Any homomorphic image of a divisible group is divisible.

(3) Finite abelian groups can not be divisible

(4) Divisible groups cannot have finite-index subgroups (this is just (2)+(3))

(5) The multiplicative group of an algebraically closed group is abelian

Now apply Jyrki's answer...


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