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.
$\blacksquare$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $12$: Applications of Sylow Theory: $(2)$ Groups of order $21$