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I know that there is a formal definition isomorphism but for the purpose of this homework questions, I call two groups isomorphic if they have the same structure, that is group table for one can be turned into the table for the other by a suitable renaming.

Now consider $\mathbf{Z}_2=\{0,1\}$ (group under addition modulo $2$) and $\mathbf{Z}_3^{\times}=\{1,2\}$ (group under multiplication modulo $3$). In general $\mathbf{Z}_n^{\times}=\{a|0\leq a\leq n-1 \text{ and }\gcd(a,n)=1\}$. Now the group table for $\mathbf{Z}_2$ is: $$ \begin{array}{c|cc} &0&1\\ \hline 0&0&1\\ 1&1&0 \end{array} $$ and the group table for $\mathbf{Z}_3^{\times}$ is: $$ \begin{array}{c|cc} &1&2\\ \hline 1&1&2\\ 2&2&1 \end{array} $$ Obviously these two groups are isomorphic because if in the first table, I replace $0$s by $1$s and $1$s by $2$s, I get exactly the second table.

However, if I write out the group table for $\mathbf{Z}_4$ and $\mathbf{Z}_5^{\times}$, there is no way that group table for one can be turned into the table for the other by a suitable renaming. This means that these two groups are not isomorphic but the question asks me to prove that they are isomorphic. Now what should I do?

  • 4
    There is no difference between the usual formal notion of "being isomorphic" and the one you propose to use!2011-04-02
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    In any case, you should look harder at the group tables for $\mathbb Z_4$ and $\mathbb Z_5^\times$!...2011-04-02
  • 1
    There is a suitable renaming. Keep trying...2011-04-02
  • 0
    In short, the two examples are all concerned with the structure of cyclic groups, et c'est la méme.2011-04-02

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