$G \in F^*$ be a finite subgroup. Show that $G$ is cyclic.

Asked Mar 28 '18 at 12:06

Active Mar 28 '18 at 12:08

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Let $F$ be a field, $F^*$ be its multiplicative group and $G \in F^*$ be a finite subgroup. Show that $G$ is cyclic.

Here is my attempt, I am not sure if it's correct.

Proof. Let $n=|G|$, we know that $1 \in G$ and $x^n=1$ for every $x \in G$. But clearly $x^{n-1} \ne 1$ for some $x \in G$, otherwise we would have $n$ distinct roots to $x^{n-1}=1$. Hence there is an $x \in G$ with order equal to $n$. $\square$

Is there an easier answer to this? And is my solution correct? Thanks.

edited Mar 28 '18 at 12:08

user1729

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asked Mar 28 '18 at 12:06

Bernoulli

2

If $x^{n-1}\neq 1$ for some $x$, then why should this $x$ have order $n$? Take $x=-1$ and $n=100$. – Dietrich Burde Mar 28 '18 at 12:08
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Have a look at the proofs at this duplicate. Then it is clear, that the proof needs more than $3$ lines. – Dietrich Burde Mar 28 '18 at 12:10
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You can also read the first two-three pages in Serre's delicious book "A course in Arithmetic" . – DonAntonio Mar 28 '18 at 12:12
thank you everyone! – Bernoulli Mar 28 '18 at 12:48

$G \in F^*$ be a finite subgroup. Show that $G$ is cyclic.

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