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Let $p$ be a prime number . The order of a $p$-Sylow subgroup of the group $GL_{50}(F_p)$ of invertible $50\times50$ matrices with entries from the finite field $F_p$,equals which of the following

1.$p^{50}$
2.$p^{125}$
3.$p^{1250}$
4.$p^{1225}$

I think order of this group is $p^{2500}$ .A sylow p-subgroup of order $P^k$ divides order of the group but $p^{k+1}$ does not. I confused here how to apply these things .

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    You might start with working out how many invertible 1x1 and 2x2 and 3x3 matrices there are - or finding out.2012-11-01
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    @Mark Bennet:please give some idea how can i find number of such matrices2012-11-01
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    Well 1x1 matrices, you want to avoid zero, and there are $p-1$ other possibilities. For 2x2 the first row can be anything but zero ($p^2-1$ possibilities) and the second row has to be linearly independent - there are $p$ multiples of the top row (including zero) to exclude, so $p^2-p$ possibilities - a total of $(p^2-1)(p^2-p)$ matrices. In such a way is the formula in Dane's answer built up for higher values of $n$.2012-11-01
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    @bdas Try accepting answers to some of your questions by clicking the green check next to your favorite answer to each question. It will motivate people to spend time on understanding your question.2012-11-01
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    @bdas Well, if you think that the group has order $p^{2500}$, and of course there are only $p^{2500}$ matrices in total over $F_p$, then it looks like you believe that *every* matrix over the field is invertible. Clearly false, right?2012-11-01
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    @rschwieb :sorry ,you are right2012-11-01

2 Answers 2

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Every element of $\operatorname{GL}_n(\mathbb{F}_p)$ is determined by the $n$ nonzero linearly independent vectors in $\mathbb{F}_p^n$ which form its columns. Counting the number of ways you can choose these reveals that the order of $\operatorname{GL}_n(\mathbb{F}_p)$ is $(p^n-1)(p^n-p)\cdots (p^n-p^{n-1})$. If $n=50$, then what is the power of $p$ dividing this order?

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The order of $p-$ sylow subgroups in $GL_n(\mathbb{F}_p)$ is $p^{n(n-1)/2}$. Put $n=50$, we get the order as $p^{1225}.$