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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

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    $\begingroup$ Amazingly, $H$ is always cyclic. And $G$ is always a direct sum of cyclics. $\endgroup$ – Gregory Grant May 19 '15 at 16:24
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    $\begingroup$ See here, for example: math.stackexchange.com/questions/837562/… $\endgroup$ – Gregory Grant May 19 '15 at 16:25
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    $\begingroup$ 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
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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, $$\mathbf F_9^{\times}\simeq \mathbf Z/8\mathbf Z$$

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

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  • $\begingroup$ 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

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