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.

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