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Let $p$ be a prime number, and $F_p$ be a field which consists of $p$ elements. With this knowledge I have to show that

  1. $k$-dimensional linear space upon the field $F_p$ has $p^k$ elements.
  2. Evaluate how many are there invertible matrices of size $n \times n$ with expressions in $F_p$ field.
  3. How many sequences (in $n$-dimensional linear space $V$) of subspaces $W_1 \subset W_2\subset \ldots W_{n-1}\subset W_n$ s. t. $\forall_{i\in \{1,2,\ldots,n\}}$ dimension of $W_i$ (upon $F_p$) is equal to $i$ are there in $V$?

I have found only something related with first question which is contained here: How many k-dimensional subspaces there are in n-dimensional vector space over $\mathbb F_p$?, but I don't see it giving me the desirable answer ($p^k$).

Any help will be much appreciated.

Novice
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    That's a harder problem than your 1. Can you count how many vectors $(a_1,\ldots,a_k)$ there are with each $a_i\in\Bbb F_p$. – Angina Seng Sep 03 '20 at 14:24
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    For 2. try to think of how to choose each row independent from the other. How many ways are there to choose the first row for an invertible matrix? Now that you have done that, how many ways to choose second row such that it is independent from the first row? – P-addict Sep 03 '20 at 15:41
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    Also in 3., $W_{n} = V$ right because $W_{n}$ would be an $n$-dimensional subspace inside $V$? – P-addict Sep 03 '20 at 15:43
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    @AnginaSeng thanks. now it is much easier. Each element of vector can be chosen in $p$ ways, and because there are $k$ elements, the result is $p^k$. – Novice Sep 04 '20 at 10:27
  • @P-addict of course first row can be chosen arbitralily, so in $p^n$ ways. Anyway, I have a problem how to choose the second in the way to make it independent from the first. Any advices will be helpful. – Novice Sep 04 '20 at 10:55
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    @Novice Now for the second row, the second row can be anything but multiples of the first row. How many multiples does the first row have, $p$ right? So, the number of ways to choose second row is $p^{n} - p$. Continuing like this, the no of ways to choose $k$-th row is $p^{n} - p^{k}$, So, the total no. of invertible matrices is ..... – P-addict Sep 04 '20 at 12:00
  • @P-addict for $k=n$ we obtain $p^n-p^n$ so is the answer $0$? – Novice Sep 07 '20 at 10:13
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    @Novice I overlooked and made a mistake while writing it : The number of ways to choose $k$-th row is $p^{n} - p^{k-1}$, you can see this in the number of ways to choose second row in particular. Also, the first row can't be choosen in $p^{n}$ ways, because you want the first row to be anything but zero, thus it is $p^{n} -1$ or $p^{n}- p^{0}$. Hope it is all clear now, sorry for the clutter. Thus, the final answer is $\prod_{k=0}^{i=n-1} p^{n} -p^{k}$ – P-addict Sep 07 '20 at 10:24

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