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?
 A: 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.
A: 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. $(\mathbb{Z}/5\mathbb{Z})^\times$.
A: As a side comment on the two positions expressed in this very old post, maybe worths to recall why both of them rightly convey the idea of isomorphicity.

A group $G$ shows its structure as soon as we allow the internal operation to fully deploy its effects. Accordingly, we can reasonably state the following:

Definition 1. The structure of a group $G$ is the set $\theta_G:=\{\theta_a, a\in G\}\subseteq \operatorname{Sym}(G)$, where $\theta_a$ is the bijection on $G$ defined by: $g\mapsto \theta_a(g):=ag$.

Here a problem arises, if we want to determine whether two groups, $G$ and $\tilde G$, "have the same structure", since in general $\operatorname{Sym}(G)\cap\operatorname{Sym}(\tilde G)=\emptyset$, and then any attempt to "compare by overlapping" the structures $\theta_G$ and $\tilde\theta_\tilde G$ is doomed to fail. We can overcome this issue "by transporting" the structure of $G$ in $\operatorname{Sym}(\tilde G)$, and see whether we can made the "transported $\theta_G$" to overlap with $\tilde\theta_{\tilde G}$. If we succeed, then we can rightly say that $G$ and $\tilde G$ are isomorphic, since we have been able to bring the structure of one onto precisely that of the other. So, with reference to the following diagram:
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\las}[1]{\kern-1.5ex\xleftarrow{\ \ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
 &  & \\
G & \ras{\space\space\space f \space\space\space}   &   \tilde G \\
\da{\theta} &  &  \da{\tilde\theta}  \\
\operatorname{Sym}(G) & \ras{\varphi^{(f)}} & \operatorname{Sym}(\tilde G) \\
\end{array}
$$
we set forth this other:

Definition 2. Two groups, $G$ and $\tilde G$, are isomorphic if there is a bijection $f\colon G\to \tilde G$ such that the above diagram commutes, namely:
$$\tilde\theta=\varphi^{(f)}\theta f^{-1}\tag 1$$
where $\varphi^{(f)}\colon \operatorname{Sym}(G)\to \operatorname{Sym}(\tilde G)$ is the "structure transporting" bijection defined by $\sigma\mapsto f\sigma f^{-1}$.

Now, as a caracterization of such an "enabling" bijection $f$, the following holds:

Claim. Two groups $G$ and $\tilde G$ are isomorphic (as per Definition 2) if and only if there is a bijection $\psi\colon \tilde G \to G$ such that:
$$\psi(\tilde a\tilde g)=\psi(\tilde a)\psi(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G \tag 2$$

Proof.
\begin{alignat}{1}
&\tilde\theta=\varphi^{(f)}\theta f^{-1} &\iff \\
&\tilde\theta_\tilde a(\tilde g)=(\varphi^{(f)}\theta f^{-1})(\tilde a)(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=(\varphi^{(f)}\theta f^{-1})(\tilde a)(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=(\varphi^{(f)}(\theta_{f^{-1}(\tilde a)}))(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=(f\theta_{f^{-1}(\tilde a)}f^{-1})(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=f(\theta_{f^{-1}(\tilde a)}(f^{-1}(\tilde g)), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&\tilde a\tilde g=f(f^{-1}(\tilde a)f^{-1}(\tilde g)), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\
&f^{-1}(\tilde a\tilde g)=f^{-1}(\tilde a)f^{-1}(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G\\
\tag 3
\end{alignat}
So, $(1)\Rightarrow (2)$, by setting $\psi:=f^{-1}$, and $(2)\Rightarrow (1)$, by setting $f:=\psi^{-1}$. $\space\space\Box$
Therefore, a bijection between two groups with the property $(2)$ is rightly called isomorphism.
