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Let $A$ be an integral domain of characteristic $2$, and let $T=\{ a \in A \ |\ \exists n \geq 1 \text{ s.t. } a^n=1 \}$ be finite. I need to prove that $T$ is a cyclic group of odd order.

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Let $R$ be the subring of $A$ generated by $T$. It has a finite generating family over $\mathbb F_2$ which is $\{1\}\cup \{t^k, t\in T, k\leq n(t)\}$ where $n(t)$ is the least integer $\geq 1$ such that $t^n = 1$; therefore it is a finite dimensional $\mathbb F_2$ vector space in particular it is finite.

Therefore it is a finite integral domain : this implies it is a field of characteristic $2$. But now we have :

A finite field of characteristic $p$ has size $p^n$ for some $n$

And

Let $K$ be a field and $G\subset K^\times$ a finite subgroup. Then $G$ is cyclic.

One easily checks that $T\subset R^\times$ and that the converse inclusion holds (by Lagrange's theorem) so $T=R^\times$ is cyclic, and has order $|R|-1 = 2^n-1$ for some $n$.

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One can even prove that $T = \{1\}$.

Indeed, let $t \in T$ and $n$ such that $t^n=1$. then $$(1+t)(1+t+t^2+\ldots+t^{n-1}) = $$ $$1+t+t^2+\ldots+t^{n-1}+$$ $$t+t^2+\ldots+t^{n-1}+1$$ $$=0$$ Now $\ t+1\ $ is a zero divisor unless $\ t=1$ or $1+t+t^2+\ldots+t^{n-1} = 0$. The latter is not the case: Let R be the subring generated by $t$, then $R$ is an algebra over the field $\mathbb{F}_2$ with (as a vector space) basis $\{1, t, t^2, \ldots, t^{n-1}\}$ so the factor under consideration has coefficients $(1,1,\ldots,1)$.

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  • $\begingroup$ That is definitiely not true. Take $A=\mathbb F_4$, then $A^\times$ is of cardinal $3$ and of course by Lagrange's theorem any $x\in A^\times$ has $x^3=1$. You haven't proved that $\{1,t...,t^{n-1}\}$ was a basis, it's just a generating set $\endgroup$ – Max Jun 22 at 13:37
  • $\begingroup$ @Max Indeed you are right. I tried to delete the answer, but I'm not allowed to do that since it was already accepted. $\endgroup$ – Marc Bogaerts Jun 22 at 13:57
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Here is another way:

Suppose $T$ is finite. Then, recall the following lemma:

Let $G$ be a finite abelian group. Then, $G$ is cyclic iff $G$ contains at most most one subgroup for each divisor of $|G|$.

In fact, the hypothesis that $G$ be abelian can be dropped. See Dummit and Foote Proposition 5 in section 6.1. Although, the proof is slightly easier when $G$ is abelian.

Let $|T| =n$ and $k \mid n$. Since $A$ is an integral domain there are at most $k$ roots of the polynomial $x^k - 1$. Hence, if there were 2 distinct subgroups of order $k$, say $H,K$, then every element of $H \cup K$ would be a root of $x^k - 1$ by lagranges theorem. But, $|H \cup K| > k$ since these subgroups are distinct. This contradicts there being at most $k$ roots of the polynomial $x^k - 1$. Thus, $T$ has at most 1 subgroup for every order dividing $n$ and by the lemma above $T$ must be cyclic. (Note: this proof is very similar to one used to show that the multiplicative group of units in a field is cyclic. This relies on the fact that a polynomial of degree $n$ over a field has at most $n$ roots in a field. This is also true in an integral domain as you can embed your domain in its fraction field.)

Now, we have shown that $T$ is cyclic, we need to show that $|T|$ is odd. If $|T|$ were even, then there would exists a subgroup of $T$ of order 2, specifically an element of order 2. Let $x \in T$ such that $|x| = 2$. Then, $x^2 = 1 \implies x^2-1 = 0 \implies (x-1)(x+1)=0$. Since $A$ is an integral domain either $x = 1$ or $x = -1$. But, $A$ has characteristic 2 so that $-1 = 1$ in $A$. Thus, $x = 1$, contrary to $|x| = 2$. As a result, $|T|$ must be odd.

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Hint: Consider the $\mathbb{F}_2$-span of $T$.

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  • $\begingroup$ Could you be more explicit? Thanks $\endgroup$ – Christa Wolf Jun 18 at 9:54
  • $\begingroup$ Note that $T$ is closed under multiplication. Hence the $\mathbb{F}_2$-span of $T$ is a finite sub-domain of $A$, hence a field $K$. $T$ is a finite multiplicative subgroup of $K^\times$, so is cyclic. $\endgroup$ – user10354138 Jun 22 at 14:08

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