Isomorphism by Cayley Table

Theorem
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structure whose underlying sets are both finite.

Then $\left({S, \circ}\right)$ and $\left({T, *}\right)$ are isomorphic :
 * A bijection $f: S \to T$ can be found such that:
 * the Cayley table of $\left({T, *}\right)$ can be generated from the Cayley table of $\left({S, \circ}\right)$ by replacing each entry of $S$ with its image under $f$.

Necessary Condition
Let $S$ and $T$ be isomorphic.

Then by definition there exists an isomorphism $f: S \to T$.

An isomorphism is a bijection by definition.

Thus the existence of the posited bijection has been demonstrated.

By the definition of set equivalence, $S$ and $T$ have the same cardinality.

Let $\left\vert{S}\right\vert = \left\vert{T}\right\vert = n$.

Let:
 * $S = \left\{ {s_1, s_2, \ldots, s_n}\right\}$

and:
 * $T = \left\{ {t_1, t_2, \ldots, t_n}\right\}$

be such that:
 * $\forall i \in \left\{ {1, 2, \ldots, n}\right\}: f \left({s_i}\right) = t_i$

Let $S \left[{j, k}\right]$ denote the entry of $S$ in row $j$ and column $k$.

Thus:
 * $S \left[{j, k}\right] = s_j \circ s_k$

for some $s_j, s_k \in S$.

We have that $f$ is an isomorphism.

Thus:
 * $f \left({S \left[{j, k}\right]}\right) = f \left({s_j}\right) * f \left({s_k}\right)$

But:
 * $f \left({S \left[{j, k}\right]}\right) = T \left[{j, k}\right]$
 * $f \left({s_j}\right) = t_j$
 * $f \left({s_k}\right) = t_k$

Thus the Cayley table of $\left({T, *}\right)$ has been generated from the Cayley table of $\left({S, \circ}\right)$ by replacing each entry of $S$ with its image under $f$.

Sufficient Condition
Let there exist a bijection $f: S \to T$ such that the Cayley table of $\left({T, *}\right)$ can be generated from the Cayley table of $\left({S, \circ}\right)$ by replacing each entry of $S$ with its image under $f$.

By the definition of set equivalence, $S$ and $T$ have the same cardinality.

Let $\left\vert{S}\right\vert = \left\vert{T}\right\vert = n$.

Let the rows and columns of the Cayley table of $\left({S, \circ}\right)$ both be numbered from $1$ to $n$.

Let the entries heading the rows and columns be denoted $s_1, s_2, \ldots, s_n$.

Let $S \left[{j, k}\right]$ denote the entry of $S$ in row $j$ and column $k$.

By the method of construction:
 * $S \left[{j, k}\right] = s_j \circ s_k$

for some $s_j, s_k \in S$.

Let $f$ be used to generate the Cayley table of $\left({T, *}\right)$ from the Cayley table of $\left({S, \circ}\right)$.

Then for all $i, j \in \left\{ {1, 2, \ldots, n}\right\}$:
 * $f \left({S \left[{j, k}\right]}\right) = T \left[{j, k}\right]$
 * $f \left({s_j}\right) = t_j$
 * $f \left({s_k}\right) = t_k$

But by definition of the Cayley table of $\left({T, *}\right)$:
 * $T \left[{j, k}\right] = t_j * t_k$

Thus:
 * $\forall j, k \in \left\{ {1, 2, \ldots, n}\right\}: f \left({s_j \circ s_k}\right) = f \left({s_j}\right) * f \left({s_k}\right)$

demonstrating that $f$ is an isomorphism.