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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:

  1. $G\cong \Bbb Z_3\times \Bbb Z_3$
  2. $G\cong\Bbb Z_9$
  3. $H\cong \Bbb Z_2\times \Bbb Z_2\times \Bbb Z_2$
  4. $G\cong \Bbb Z_3\times \Bbb Z_3$, $H\cong\Bbb Z_8$

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.

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    $\begingroup$ $ G $ is only the additive group of your field. It does not have any multiplicative structure - saying that $ G $ has zero divisors doesn't make sense. $\endgroup$ – Starfall Oct 8 '16 at 16:04
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    $\begingroup$ For 1, the field of 9 elements has characteristic 3. By the structure theorem for abelian groups, the only possibility is $\mathbb{Z}_3\times \mathbb{Z}_3$. $\endgroup$ – Sungjin Kim Oct 8 '16 at 16:33
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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.

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  • $\begingroup$ Okay ,But How do you define $a*b$ in $\Bbb Z_2\times Z_2$ is still not clear;Please specify the rule if it is not componentwise multiplication $\endgroup$ – Learnmore Oct 9 '16 at 2:29
  • $\begingroup$ I think it's probably not useful to try to explain it here, but a short description is: You interpret $(a,b)$ as the polynomial $ax+b$. To multiply $(a,b)$ and $(c,d)$ you multiply the polynomials and then divide by $x^2+x+1$ and take the remainder. I suggest that you read your textbook. As I said, it will explain. $\endgroup$ – MJD Oct 9 '16 at 12:10
  • $\begingroup$ I don't know why to divide it by $x^2+x+1$ neither does my textbook explain it ;Please suggest some links/references $\endgroup$ – Learnmore Oct 9 '16 at 14:01
  • $\begingroup$ $x^2+x+1$ is irreducible over $\Bbb Z_2[x]$. $\endgroup$ – MJD Oct 9 '16 at 14:40

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