Let us consider the multiplication operation, denoted by $ \odot $ on the set of 2-bit integers ${Z_4}$ defined as follows:
$$\eqalign{ & a \odot b = (ab\,\bmod \,5)\,\bmod \,4\,if\,a \ne 0,\,b \ne 0 \cr & 0 \odot a = a \odot 0 = (4a\,\bmod \,5)\,\bmod \,4 \cr & 0 \odot 0 = 1 \cr} $$
The task is
- Compute the Cayley table for $ \odot $
- Show that $({Z_4}, \odot )$ is isomorphic with multiplicative group of the field ${Z_5}$
For the first part i have constructed the Cayley table. Is it correct?
\begin{array}{ccc} \odot & \textbf{0} & \textbf{1} & \textbf{2} & \textbf{3} \\ \textbf{0} & 1 & 0 & 3 & 2 & & \\ \textbf{1} & 0 & 1 & 2 & 3 & & \\ \textbf{2} & 3 & 2 & 0 & 1 & & \\ \textbf{3} & 2 & 3 & 1 & 0 & & \end{array}
How can i show that ${Z_4}$ is isomorphic with multiplicative group of the field ${Z_5}$?
Your multiplication table is correct. As for showing that your group is isomorphic to the multiplicative group $\Bbb Z_5$, note that the multiplication table is given by
\begin{array}{c|cccc} & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{array}
Note how similar this is to your multiplication table. You want a function to map $\{0,1,2,3\}$ to $\{1,2,3,4\}$ such that your multiplication table looks like mine. Isomorphic groups have identical multiplication tables (modulo "naming" of variables and rearranging of rows/columns accordingly). This is, in fact, why we call them isomorphic groups - up to naming, we can't really distinguish them apart since they have the same action. Hint: notice that $1\odot a = a = a\odot 1$ in your group and look at the diagonals in our multiplication tables - these will give you a couple of good ideas as to how to map things.