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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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Tried to fix your TeX code. Does this looks good? $$\left\{ \left(\begin{array}{ccc} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{array}\right)‎‎\vert‎‎‎a,b,c\in‎\mathbb{F}_{3}\right\} ‎‎\cong(‎\mathbb{Z}_{3}‎\times\mathbb{Z}_{3})‎\rtimes\mathbb{Z}_{3}$$ –  zaarcis Feb 10 '13 at 22:09
    
Yes of course! That's right. many thanks Ilmārs Cīrulis! I tried so much to write this but I couldn't. –  shankfei Feb 10 '13 at 22:13
2  
How is it that the fact is well-known and you are asking these questions?! :-) –  Mariano Suárez-Alvarez Feb 10 '13 at 22:17
    
It's nice. :) But I messed little bit up lines (= lost few linebreaks) and can't fix it because such change is too small to be allowed. –  zaarcis Feb 10 '13 at 22:19
    
@ Mariano Suárez-Alvarez! actually I think it is well-known, because I saw it in an article without any detail. Maybe it isn't! –  shankfei Feb 10 '13 at 22:25

2 Answers 2

up vote 3 down vote accepted

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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Dear Andreas Caranti thank you! –  shankfei Feb 10 '13 at 22:30
    
@shankfei, you're welcome! –  Andreas Caranti Feb 10 '13 at 22:32

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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