# Abstract algebra(group theory) - alternative proof (property of general linear group)

Let: $n\in\mathbb{N}$, $p$ - prime, $Z_{p} :=\{0,1,\dots,p-1\}$, $GL_{n}(Z_{p})$ - general linear group of order $n$ whose elements of its matrices are in $Z_{p}$. Prove: $|GL_{n}(Z_{p}|=\prod\limits_{k=0}^{n-1}(p^{n}-p^{k})$

Thought of proving this proposition by the use of induction with respect to $n$, but failed to derive constructive proof(i.e. not relying on principle of mathematical induction). If anyone would be able to post constructive proof, then I would be very thankful.

• I don't see why you consider the inductive proof to be "non-constructive"! The inductive proof essentially constructs the matrices, row by row. See this answer: math.stackexchange.com/questions/2164601/…. Commented Mar 1, 2017 at 16:39
• @Kenny Wong Induction assumes existence of conjecture at the beginning of the proof. I know proof of the above proposition using principle of mathematical induction. I only wonder how to construct proof "from scratch" assuming that conjecture is unknown.
– user410985
Commented Mar 1, 2017 at 16:46

Let us look at a general element in $\;GL_n(\Bbb Z_p)\;$, and at its columns as vectors in $\;\left(\Bbb Z_p\right)^n\;$ : in how many ways you can choose the first column? Well, you can take any element in $\;\left(\Bbb Z_p\right)^n\;$ but one (can you see which one?), so in $\;p^n-1\;$ ways.
Now, in how many ways can you choose the second column? Any element in $\;\left(\Bbb Z_p\right)^n\;$ will work except the scalar multiples of the first one we already chose , so $\;p^n-p\;$ possible vectors.