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By a well known fact we have: \begin{equation} \left\{ \left(\begin{array}{ccc} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{array}\right)\verta,b,c\in\mathbb{F}_{3}\right\} \cong(\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{3} \end{equation} Now I have two questions: 1) What is the action of $\mathbb{Z}_3$ on $\mathbb{Z}_3 \times \mathbb{Z}_3$? 2) Under which isomorphism these two groups are isomorphic? Many thanks. |
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You may take for $\mathbb{Z}_3\times \mathbb{Z}_3$ the normal subgroup $$ \left\{ \alpha_{a,b} = \begin{bmatrix} 1 & a & b\\ & 1 & \\ &&1 \end{bmatrix} : a, b \in \mathbb{F}_{3} \right\}, $$ (I am omitting zeroes) and for the "other" $\mathbb{Z}_3$ the subgroup $$ \left\{ \gamma_{c} =\begin{bmatrix} 1 & & \\ & 1 & c \\ &&1 \end{bmatrix} : c \in \mathbb{F}_{3} \right\}. $$ It is easly checked that the action is given by $$ \gamma_{c}^{-1} \alpha_{a,b} \gamma_{c} = \alpha_{a, ac+b}. $$ This should tell you everything you want. |
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Write the matrix in your question as $(a,b,c)$ for simplicity and —computing explicitly— find the formula for the product $(a,b,c)\cdot(a',b',c')$ This should make it obvious how the semi-direct product is constructed. |
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