Isomorphism in a field Let $(F,+,.)$ is the finite field with $9$ elements. Let $G=(F,+)$ and $H=(F\setminus\{0\},\cdot)$ denote the underlying additive and multiplicative groups. Then:


*

*$G\cong \Bbb Z_3\times \Bbb Z_3$

*$G\cong\Bbb Z_9$

*$H\cong \Bbb Z_2\times \Bbb Z_2\times \Bbb Z_2$

*$G\cong \Bbb Z_3\times \Bbb Z_3$, $H\cong\Bbb Z_8$


Since $F$ is a field thus $H$ is cyclic so $H\cong\Bbb Z_8$.
Now $\Bbb Z_3\times \Bbb Z_3$ has zero divisors since $(1,0)\times (0,1)=(0,0)$ and a field can't have zero divisors.
Also $\Bbb Z_9$ also has zero divisors since $[3]\times [3]=[0]$ and a field can't have zero divisors.
So none of the options are correct.
But answer is 1,4 are correct. Please help.

Say $H$ is a field with $p^k$ elements.  Then you can understand $\langle H, +\rangle$ as being $\Bbb Z_p^k$, where the elements are $k$-tuples of elements from $\Bbb Z_p$, and the addition operation is component-wise addition in $\Bbb Z_p$.  But if you do this, the multiplication for $H$ is not component-wise multiplication of the $k$-tuples.
For example, consider the field $H$ with $2^2 = 4$ elements:
$$
\begin{array}{c|cccc}
+ & e & a & b & c \\ \hline
e & e & a & b & c \\
a & a & e & c & b \\
b & b & c & e & a \\
c & c & b & a & e                                                                          \end{array}\qquad
\begin{array}{c|cccc}
× & e & a & b & c \\ \hline                                                                                                                                                                                        
e & e & e & e & e \\                                                                                                                                                                                               
a & e & a & b & c \\                                                                                                                                                                                               
b & e & b & c & a \\                                                                                                                                                                                               
c & e & c & a & b    \end{array}
$$
There is only one field with 4 elements, and its operation tables must look like this.
We can interpret the elements $e,a,b,c$ as being pairs of elements from $\Bbb Z_2$, respectively $(0,0), (0,1), (1,0),$ and $(1,1)$.  If we do that, and replace $e$ with $(0,0)$ and so on in the tables above, they look like this:
$$
\begin{array}{c|cccc}
+ & (0,0) & (0,1) & (1,0) & (1,1) \\ \hline
(0,0) & (0,0) & (0,1) & (1,0) & (1,1) \\
(0,1) & (0,1) & (0,0) & (1,1) & (1,0) \\
(1,0) & (1,0) & (1,1) & (0,0) & (0,1) \\
(1,1) & (1,1) & (1,0) & (0,1) & (0,0)
\end{array}\qquad
\begin{array}{c|cccc}
× & (0,0) & (0,1) & (1,0) & (1,1) \\ \hline
(0,0) & (0,0) & (0,0) & (0,0) & (0,0) \\
(0,1) & (0,0) & (0,1) & (1,0) & (1,1) \\
(1,0) & (0,0) & (1,0) & (1,1) & (0,1) \\
(1,1) & (0,0) & (1,1) & (0,1) & (1,0)
\end{array}
$$
This can help you understand the way addition works, because the left-hand table, for addition, is exactly the table for component-wise addition of pairs of mod-2 integers: $(1,1) + (1,0) = (0,1)$ for example.
But you cannot understand the multiplication this way, because the right-hand table is not component-wise multiplication of pairs of mod-2 integers; it has $(1,1)\times (1,0) = (0,1)$ rather than $(1,0)$.
To understand both tables at once, you need to do something more interesting, such as interpreting the elements as certain classes of polynomials with coefficients in $\Bbb Z_p$, and your algebra class will probably teach you to do that, if it has not already.
