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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

Jesse P Francis
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2 Answers2

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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, $$\mathbf F_9^{\times}\simeq \mathbf Z/8\mathbf Z$$

Bernard
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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.

lhf
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