Groups of Order 21/Matrix Representation of Non-Abelian Instance

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Theorem

Let $G$ be the group of order $21$ whose group presentation is:

$\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2}$


Then $G$ can be instantiated by the following pair of matrices over $\Z_7$:

$X = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \qquad Y = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}$


Proof

We calculate the powers of $X$ and $Y$ in turn:

\(\ds X^2\) \(=\) \(\ds \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\)
\(\ds X^3\) \(=\) \(\ds \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}\)

and so on to:

\(\ds X^7\) \(=\) \(\ds \begin{pmatrix} 1 & 6 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) as $7 \equiv 0 \pmod 7$


Thus we have:

$X^7 = \mathbf I$

where $\mathbf I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the identity matrix.


Then:

\(\ds Y^2\) \(=\) \(\ds \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}\)
\(\ds Y^3\) \(=\) \(\ds \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)


Thus we have:

$Y^3 = \mathbf I$

and:

$Y^{-1} = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$.


Then:

\(\ds Y X Y^{-1}\) \(=\) \(\ds \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \begin{pmatrix} 4 & 4 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}\)
\(\ds \) \(=\) \(\ds \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\)
\(\ds \) \(=\) \(\ds X^2\)

and the result is apparent.

$\blacksquare$


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