Let $(F,+,\cdot)$ is the finite field with $9$ elements. Then which of the following are true? Let $(F,+,\cdot)$ is the finite field with $9$ elements. Let $G=(F,+)$ and $H=(F-\{0\}, \cdot)$ denote the underlying additive and multiplicative groups respectively. Then, 


*

*$G\cong (\mathbb{Z/3Z})\times(\mathbb{Z/3Z})$

*$G\cong\mathbb{Z/9Z}$

*$H\cong(\mathbb{Z/2Z})\times(\mathbb{Z/2Z})\times(\mathbb{Z/2Z})$

*$G\cong (\mathbb{Z/3Z})\times(\mathbb{Z/3Z})$ and $H\cong\mathbb{Z/8Z}$


I guess $1$ and $2$ and $4$ are true but not convincingly. Can anybody help me out? Hints will be fine.
 A: $G$ can be seen as a vector space over its subfield $\{0,1,2\}$, of dimension 2; this is the usual argument that a field has a prime power as its cardinality.
This shows 1) holds. This rules out 2), as $\mathbb{Z} / 9\mathbb{Z}$ has an additive element of order 9, while the first one has not.
For any field $F$, $F \setminus \{0\}$ is cyclic (as a multiplicative group) (theorem), which rules out 3. Check that 4 is indeed true.
A: Note that the characteristic of the field is 3.So the field F is the splitting field of  $x^{3^2}-x$ on $F_3$. I doubt whether (1) holds. Since (0,1) does not have a reverse.
Update: Since G is an ADDITIVE group I think (1) holds.
A: For (3) and (4) if you want a elementary reasoning, you can do this:
In $H$ any element has order $1,2,4$ or $8$. If you assume by contradiction that there is no element of order $8$, then all $8$ elements of $H$ satisfy 
$$x^4=1$$
Therefore this polynomial has 8 roots, contradiction.
This shows that $H$ has an element of order $8$ and is thus cyclic.
P.S. This is the basic idea behind the proof of the Theorem quoted in the other answer.
For (1) or (2) We show that every element $x \neq 0$ in $G$ must have order $3$:
In $F$ you have
$$(x+x+x) \cdot (x+x+x) = 9x^2 =0$$
with the last equality following from the fact that $G$ has order $9$.
Since $F$ is a field you get $$x+x+x=0$$
Now, since $G$ is an abelian group with $9$ elements, it is easy to prove that the above relation implies $G \sim (\mathbb{Z/3Z})\times(\mathbb{Z/3Z})$, as there are only 2 abelian groups with 9 elements.
If you want a Group Theoretical proof, pick $x \neq 0$ in $G$ and $y \in G \backslash \{ 0, x, 2x \}$. It is very easy to show that 
$$G \sim < x> \times <y> $$ 
A: Suppose $B$ is a matrix on $F_3$
 $$B = \begin {bmatrix} {0} & {1}\\ {1} & {2} \end{bmatrix}
$$   $$F_9 = \{ {0}, {I} , {B} ,{B^{2}} ,{B^{3}} ,{B^{4}} ,{B^{5}} ,{B^{6}} ,{B^{7}}  \} $$
then $NO.4$ is correct. because two finite fields with the same number of elements are isomorphic .
