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Problem of the Week #276 - Aug 15, 2018

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Euge

MHB Global Moderator
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Jun 20, 2014
1,894
Here is this week's POTW:

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For $p$ a prime integer, compute the order of the special linear group $SL_n(\Bbb F_p)$, where $\Bbb F_p$ is the field with $p$ elements.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,894
This week’s problem was solved correctly by Olinguito . You can read his solution below.


We first compute the order of the general linear group $\mathrm{GL}_n(\mathbb F_p)$.

The first column can be any $n\times1$ column vector except the zero vector and so there are $p^n-1$ choices for the first column. The second column can be any vector except a linear multiple of the first, giving $p^n-p$ choices for the second column. In general, the $i$th column can be any column vector except a linear combination of the first $i-1$; there are $p^{i-1}$ such linear combinations and so there are $p^n-p^{i-1}$ choices for the $i$th column. Hence we get $$\left|\mathrm{GL}_n(\mathbb F_p)\right|\ =\ (p^n-1)(p^n-p)\cdots(p^n-p^{n-1}).$$
Next, consider the determinant function $\det:\mathrm{GL}_n(\mathbb F_p)\to\mathbb F_p^\times$, the multiplicative group of the nonzero elements of $\mathbb F_p$. This function is surjective, for, given $a\in\mathbb F_p^\times$, if $A$ is the diagonal matrix with $(1,1)$th entry $a$ and all other entries on the leading diagonal $1$, then $\det(A)=a$. It is also a homomorphism by the multiplicative property of determinants. The kernel of this homomorphism is the group of all matrices with determinant $1$, namely $\mathrm{SL}_n(\mathbb F_p)$. Hence, we have
$$\mathrm{GL}_n(\mathbb F_p)/\mathrm{SL}_n(\mathbb F_p)\ \cong\ \mathbb F_p^\times$$
and so$$\left|\mathrm{SL}_n\left(\mathbb F_p\right)\right|\ =\ \frac{\left|\mathrm{GL}_n\left(\mathbb F_p\right)\right|}{\left|\mathbb F_p^\times\right|}\ =\ \frac{\displaystyle\prod_{i=0}^{n-1}\left(p^n-p^i\right)}{p-1}.$$
 
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