Suppose $F$ is any field. Let $G$ and $H$ are two subgroups of $F^*$ of order $n$. Then is it true that $G=H$?

For finite field this is true. Is this true for arbitrary fields?


Yes, this is true. For every $a\in G$, $a^n=1$. So every $a\in G$ is a root of the polynomial $$ f(x)=x^n-1\in F[x]. $$ The same holds for every $b\in H$. Because $|G|=|H|=n$, $G$ and $H$ consist of all the roots of $f(x)$ and so $G=H$.

| cite | improve this answer | |
  • 1
    $\begingroup$ Note that this implicitly uses the fact that $x^n-1$ has at most $n$ roots in $F$. $\endgroup$ – Servaes Sep 27 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.