My Math Forum solvable Group of matrices over Z_p

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 November 4th, 2013, 02:03 AM #1 Newbie   Joined: Nov 2013 Posts: 26 Thanks: 0 solvable Group of matrices over Z_p show that the set of invertible upper triangular $2\times 2$ matrices over $Z_p$ is a Group for matrix-multiplication.\\ so I showed that the set $M :=\left\{ \left(\begin{array}{cc} a&b\\ 0=&c \end{array} \right) | a,c\ne 0 ; a,b,c \in Z_p \right\}=$ is a group by showing the group-axioms. And the order of the Group is then $(1-p)^2\cdot p$ \item show that the set $N :=\left\{ \left(\begin{array}{cc} 1&b\\ 0=&1 \end{array} \right) | b \in Z_p \right\}=$ is a commutative normal subgroup of M .\\ that was also no problem ( just checking the subgroup-criterium+ normality) \item ... but now comes my Problem : show that M is solvable by using this normal subgroup:\\ \vspace{0.5cm} I get the series of normal subgroups: $\{id\} \triangleleft N \triangleleft M$ with the factors$F_1 \simeq Z_p$ and $F_2= M/N$.\\ $F_1$ is cyclic and so abelian. but my problem is $F_2$ ... if i could show that $F_2$ is abelian , then M would be solvable ... but how can i get that???? \end{enumerate}[/latex]
 November 4th, 2013, 04:19 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: solvable Group of matrices over Z_p Actually, a general result holds true : The group of invertible upper-triangular martices over any finite field is solvable. Try showing the finiteness of the derived series of M.
 November 4th, 2013, 04:58 AM #3 Newbie   Joined: Nov 2013 Posts: 26 Thanks: 0 Re: solvable Group of matrices over Z_p But if the Order of M = $(p-1)^2\cdot p$ then this series has to be finite ... or am i completely on the wrong way ? I wanted to show that with a theorem shown in our course : Let $N \triangleleft M$. If N and M/N are solveable, then M is solveable. N is of course solveable because it was order = p . But M/N would have order = $(p-1)^2$ and how can i show that this is solveable ..
November 4th, 2013, 06:14 AM   #4
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Re: solvable Group of matrices over Z_p

Quote:
 Originally Posted by Sandra93 But if the Order of M = $(p-1)^2\cdot p$ then this series has to be finite
Why do you think so? (NOTE* : I have not checked your calculation of order, but I think it is correct).

Quote:
 Originally Posted by Sandra93 N is of course solvable because it was order = p.
Is p a prime?

Quote:
 Originally Posted by Sandra93 But M/N would have order = $(p-1)^2$ and how can i show that this is solveable ..
I am not sure how to use the order to prove solvability of M/N. In fact, I am unsure if it can be approached this way. My argument using induction seems much more better, and I am not sure why you think that is unsatisfactory.

 November 4th, 2013, 09:43 AM #5 Newbie   Joined: Nov 2013 Posts: 26 Thanks: 0 Re: solvable Group of matrices over Z_p oh i forgot this part .. yes p is a prime. and the order is $p\cdot (p-1)^2$ because i have $p-1$ possibilities for the two elements on the diagonal, and $p$ possibible values for the element right on the top ... And i donīt think that your way is unsatisfactory. To be honest... i just didnīt get what you meaned by "showing the finiteness"
November 4th, 2013, 10:58 AM   #6
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Re: solvable Group of matrices over Z_p

Quote:
 Originally Posted by Sandra93 oh i forgot this part .. yes p is a prime.
Then indeed N is solvable, by Feit-Thompson theorem.

Quote:
 Originally Posted by Sandra93 i just didnīt get what you meaned by "showing the finiteness"
I meant that the derived series is finite. This can be done by calculating commutator subgroups respectively. Try computing G1 = [G, G]. Do you notice any pattern for the elements of G1?

 November 4th, 2013, 11:53 AM #7 Newbie   Joined: Nov 2013 Posts: 26 Thanks: 0 Re: solvable Group of matrices over Z_p ok I'll try it: Let $A=\left( \begin{array}{cc} a_1 &a_2 \\ 0=&a_3 \end{array}\right)=$ and $B=\left( \begin{array}{cc} b_1 &b_2 \\ 0=&b_3 \end{array}\right)=$ be in M: $[A,B]= A^{-1}B^{-1}AB$ ... and now i just try to multiplate this 4 matrices: $=\left( \begin{array}{cc} a_1^{-1} & -a_2(a_1a_3)^{-1} \\ 0 & a_3^{-1} \end{array}\right)\left( \begin{array}{cc} b_1^{-1} & -b_2(b_1b_3)^{-1} \\ 0 & b_3^{-1} \end{array}\right)\left( \begin{array}{cc} a_1 & a_2 \\ 0 & a_3 \end{array}\right)\left( \begin{array}{cc} b_1 & b_2 \\ 0 & b_3 \end{array}\right)$ = $\left( \begin{array}{cc} a_1^{-1}b_1^{-1} & -a_1^{-1}b_2(b_1b_3)^{-1}-b_3^{-1}a_2^{-1}(a_1a_3)^{-1} \\ 0 & a_3^{-1}b_3^{-1} \end{array}\right)\cdot \left( \begin{array}{cc} a_1b_1 & a_1b_2+a_2b_3 \\ 0 & a_3b_3 \end{array}\right)$ $=\left( \begin{array}{cc} 1 & (**) \\ 0 & 1 \end{array}\right)$ ... where (**) is a giant term , but definitly element of $Z_p$ is this, what you meaned?? and the pattern is that the elements have the same form als the elements in N?
November 4th, 2013, 12:05 PM   #8
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Re: solvable Group of matrices over Z_p

Quote:
 Originally Posted by Sandra93 is this, what you meaned?? and the pattern is that the elements have the same form als the elements in N?
Not quite. Note that the elements of [M, M] has all 1s in the main diagonal. Now use this fact to show that elements of [[M, M], [M, M]] consists only of the identity element of M. This can be done by taking elements of N, say, X and Y and then calculating [X, Y].

PS : I would think that it is also provable that [M, M] is indeed N, but that'd be harder the other way around.

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