Algebra Solutions: A finite group is abelian if and only if its Cayley table is a symmetric matrix

Prove that a finite group is abelian if and only if its group table is a symmetric matrix.

If $G$ is a finite group, there exists a bijective indexing map $\varphi : [1,n] \rightarrow G$ for some natural number $n$ ; we can then define the group table $T = [t_{i,j}]$ of $G$ induced by $\varphi$ to be a matrix with $t_{i,j} = g_i \cdot g_j$ .
$(\Rightarrow)$ Suppose $G$ is a finite abelian group. Then for all $1 \leq i,j \leq n$ we have $t_{i,j} = g_i \cdot g_j = g_j \cdot g_i = t_{j,i}$ . Hence $T$ is symmetric.
$(\Leftarrow)$ Suppose $T$ is a symmetric matrix. Then for all $1 \leq i,j \leq n$ we have $g_i \cdot g_j = t_{i,j} = t_{j,i} = g_j \cdot g_i$ , so that $G$ is abelian.

Algebra Solutions

A finite group is abelian if and only if its Cayley table is a symmetric matrix

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