Identity Elements occupy Diagonal of Cayley Table in Inverse Row Form

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Theorem

Let $\left({G, \circ}\right)$ be a finite group.

Let $\mathcal C$ be a Cayley table for $\left({G, \circ}\right)$ presented in inverse row form.

Then all the entries in the main diagonal of $\mathcal C$ are instances of the identity element.


Proof

By definition of inverse row form, the rows of $\mathcal C$ are headed by the inverse elements of the elements which head the corresponding columns.

The entries in the main diagonal of $\mathcal C$ have the same column number as row number.

Let $\left[{c}\right]_{k k}$ denote the entry of $\mathcal C$ corresponding to the element where the $k$th row intersects the $k$th column.

Let $a$ be the element which heads column $k$.

Then, by definition, $a^{-1}$ is the element which heads row $k$.

Thus the element which occupies entry $\left[{c}\right]_{k k}$ is $a^{-1} \circ a = e$.

Hence the result.

$\blacksquare$


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