**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.10**

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

Solution: 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.