Exercise 1.1.10 (Group table of a finite abelian group)

Question

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

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $G = \{a_1, a_2, \ldots,a_n\}$ be a finite group under multiplication, and let $M = (m_{ij})_{1 \leq i \leq n, 1 \leq j \leq n} $ be the matrix defined by $m_{ij} = a_i \cdot a_j$. Then
\begin{align*}
G 	ext{ is abelian }& \iff \forall i \in [\![1, n]\!], \ \forall j\in [\![1, n]\!], \ a_i \cdot a_j = a_j \cdot a_i\
&\iff  \forall i \in [\![1, n]\!], \ \forall j\in [\![1, n]\!], \ m_{ij} = m_{ji}\
&\iff M = M^{t}.
\end{align*}
So $G$ is abelian if and only if its group table is a symmetric matrix.
\end{proof}

Exercise 1.1.10 (Group table of a finite abelian group)

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Exercise 1.1.10 (Group table of a finite abelian group)

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