Let $(F,+,.)$ be a finite field with 9 elements. Let $G=(F,+)$ and $H=(F\setminus \{0\},.)$ denote the underlying additive and multiplicative groups. Then what will $G$ and $H$ be isomorphic to?

We know that any finte abelian group is a direct product of cyclic group thus either $G$ is isomorphic to $\mathbb Z_9$ or $\mathbb Z_3\times \mathbb Z_3$ and $H$ is isomorphic to either $\mathbb Z_8$ or $\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_2$. Since a field can have no zero divisor theus $G$ willbe isomorphic to $\mathbb Z_3\times \mathbb Z_3$

But I can't conclude what $H$ will be isomorphic to. Any help

  • 2
    $\begingroup$ Amazingly, $H$ is always cyclic. And $G$ is always a direct sum of cyclics. $\endgroup$ – Gregory Grant May 19 '15 at 16:24
  • 1
    $\begingroup$ See here, for example: math.stackexchange.com/questions/837562/… $\endgroup$ – Gregory Grant May 19 '15 at 16:25
  • 1
    $\begingroup$ What I'm trying to say is $H$ is cyclic no matter what $F$ is, as long as it is a finite field. Most people find that fact pretty startling. $\endgroup$ – Gregory Grant May 19 '15 at 16:26
  • $\begingroup$ In principle, $H$ could also be isomorphic to $\mathbb Z_2\times \mathbb Z_4$. $\endgroup$ – lhf May 19 '15 at 16:34

Actually any finite subgroup of the multiplicative group of a field (whether the field itself is finite or not) is cyclic. In the present case, $$\mathbf F_9^{\times}\simeq \mathbf Z/8\mathbf Z$$

| cite | improve this answer | |
  • $\begingroup$ Asimple question sir ...Why? $\endgroup$ – Learnmore May 19 '15 at 17:17
  • 1
    $\begingroup$ @learnmore: see my answer to this question. $\endgroup$ – Bernard May 19 '15 at 17:43

Here is an elementary argument that does not even need the structure theorem of abelian groups:

Let $n$ be the exponent of $H$, that is, the smallest $n$ such that $h^n=1$ for all $h\in H$.

By Lagrange's theorem, $n\le 8$.

If $n<8$, then equation $x^n=1$ would have $8$ solutions, and this cannot happen in a field.

Hence, $n=8$.

If all elements have order less than $8$, then the exponent is at most $4$, because the possible orders are $1$, $2$, or $4$.

Since the exponent is $8$, $H$ must then have an element of order $8$ and so $H$ is cyclic.

| cite | improve this answer | |
  • $\begingroup$ what is exponent? $\endgroup$ – Learnmore May 19 '15 at 17:16
  • $\begingroup$ @learnmore, see my edited answer. $\endgroup$ – lhf May 19 '15 at 17:33

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.