# Isomorphic groups

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?

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There is no difference between the usual formal notion of "being isomorphic" and the one you propose to use! – Mariano Suárez-Alvarez Apr 2 '11 at 5:10
In any case, you should look harder at the group tables for $\mathbb Z_4$ and $\mathbb Z_5^\times$!... – Mariano Suárez-Alvarez Apr 2 '11 at 5:10
There is a suitable renaming. Keep trying... – Zev Chonoles Apr 2 '11 at 5:12
In short, the two examples are all concerned with the structure of cyclic groups, et c'est la méme. – awllower Apr 2 '11 at 7:08

You are incorrect in claiming that there is no way to turn one table into the other. But the key is that you are not just allowed to rename, you are also allowed to list the elements in a different order! (After all, listing the elements of $\mathbf{Z}_4$ in a different order in the table will not change the group or the operation, will it?) This amounts to shuffling rows and columns together: if you exchange rows 2 and 3, say, you should also exchange columns 2 and 3.

The table for $\mathbf{Z}_4$ is: $$\begin{array}{c|cccc} +&0&1&2&3\\ \hline 0&0&1&2&3\\ 1&1&2&3&0\\ 2&2&3&0&1\\ 3&3&0&1&2 \end{array}$$ The table for $Z_5^{\times}$ is: $$\begin{array}{c|cccc} \times&1&2&3&4\\ \hline 1&1&2&3&4\\ 2&2&4&1&3\\ 3&3&1&4&2\\ 4&4&3&2&1 \end{array}$$ If you are going to be able to rename the entries in the last table to match the first, then "1" must be renamed "0". Now, notice that there is only one of the remaining four elements that when operated with itself gives you the "identity"; since in $\mathbf{Z}_4$ this happens for $2$, you may want to shuffle the rows and columns to move that element to be in the third row and column and see what you have then.

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c'est superbe!! – awllower Apr 2 '11 at 7:11
Apparently not; any reason for the downvote? – Arturo Magidin Apr 2 '11 at 18:13
Who downvoted ? – awllower Apr 3 '11 at 2:45
Don't know; but someone did... – Arturo Magidin Apr 3 '11 at 2:50
Since $\mathbb{Z}_n^\times$ is cyclic, reordering can be easily done by picking a generator element (i. e., any other than 1), and computing its power. In this case, the powers of 2 are, in order, 1, 2, 4 and 3, which is the ordering for the first line of the table for $\mathbb{Z}_5^\times$ that will suit the isomorphism. – Luke May 10 '11 at 1:04

Your "naive" definition of isomorphic almost coincides with the abstract one. The difference is that apart from renaming you must also be allowed to change the order of the elements. Then you should have no difficulties with $\mathbb{Z}/4\mathbb{Z}$ vs. $(\mathbb{Z}/5\mathbb{Z})^\times$.

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Note by the way that this little confusion is a good argument for learning the actual definition of group isomorphism, rather than trying to avoid it by only thinking in terms of "group tables". – Pete L. Clark Apr 2 '11 at 12:30