# Prove by induction that $R$ (subring of matrices $3\times3$ over Galois field) satisfies $(xy)^{k}=x^{k}y^{k}$

I need to prove by induction that $R$, which is subring of matrices $3\times3$ over Galois field defined by

$$R=\Biggl\{\begin{pmatrix}a & b & c\\ 0 & a & d\\ 0 & 0 & a \end{pmatrix}\, a,\, b,\, c,\, d\in GF(p)\Biggr\}$$

satisfies $(xy)^{k}=x^{k}y^{k}$ for every $x, y\in R$ and $k>2$.

I proved that this is true for $k=3$ and claiming that it is true for $k$ I get that

$\Biggl\{\begin{pmatrix}a & b & c\\ 0 & a & d\\ 0 & 0 & a \end{pmatrix}\begin{pmatrix}e & f & g\\ 0 & e & h\\ 0 & 0 & e \end{pmatrix}\Biggr\}^{k+1} = \Biggl\{\begin{pmatrix}a & b & c\\ 0 & a & d\\ 0 & 0 & a \end{pmatrix}\begin{pmatrix}e & f & g\\ 0 & e & h\\ 0 & 0 & e \end{pmatrix}\Biggr\}^{k}\begin{pmatrix}a & b & c\\ 0 & a & d\\ 0 & 0 & a \end{pmatrix}\begin{pmatrix}e & f & g\\ 0 & e & h\\ 0 & 0 & e \end{pmatrix} = \begin{pmatrix}a & b & c\\ 0 & a & d\\ 0 & 0 & a \end{pmatrix}^{k}\begin{pmatrix}e & f & g\\ 0 & e & h\\ 0 & 0 & e \end{pmatrix}^{k}\begin{pmatrix}a & b & c\\ 0 & a & d\\ 0 & 0 & a \end{pmatrix}\begin{pmatrix}e & f & g\\ 0 & e & h\\ 0 & 0 & e \end{pmatrix}$

However, I don't know how to continue because this ring isn't commutative. Could anyone help me to finish this proof by induction, please? Thank you.

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$$\begin{pmatrix}{}1&1&0\\0&1&0\\0&0&1\end{pmatrix}^3 \begin{pmatrix}{}1&0&0\\0&1&1\\0&0&1\end{pmatrix}^3 = \begin{pmatrix}{}1&3&0\\0&1&0\\0&0&1\end{pmatrix} \begin{pmatrix}{}1&0&0\\0&1&3\\0&0&1\end{pmatrix} = \begin{pmatrix}{}1&3&9\\0&1&3\\0&0&1\end{pmatrix} \\ \left(\begin{pmatrix}{}1&1&0\\0&1&0\\0&0&1\end{pmatrix} \begin{pmatrix}{}1&0&0\\0&1&1\\0&0&1\end{pmatrix}\right)^3 = \begin{pmatrix}{}1&1&1\\0&1&1\\0&0&1\end{pmatrix}^3 = \begin{pmatrix}{}1&3&6\\0&1&3\\0&0&1\end{pmatrix}\\$$
Ok, perhaps we should assume $p=3$? –  Martin Brandenburg Sep 18 '13 at 10:56
@MartinBrandenburg : then we can just pick $k=5$ instead of $3$. With general $k$, the topright coefficients are $k^2$ and $k(k+1)/2$, their difference is $k(k-1)/2$ and there is no prime $p$ dividing all of those numbers. –  mercio Sep 18 '13 at 15:30