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I have $SL_{2}\{\mathbb{Z}/p\}$ for $p$ prime and $\mathbb{Z}$ integers. How do I show that this is a subgroup of $GL_{2}\{\mathbb{Z}/p\}$ and find the number of elements in $SL_{2}\{\mathbb{Z}/p\}$?

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  • $\begingroup$ The subgroup part is easy, just show the identity is in $SL_2(\Bbb Z/p\Bbb Z)$ and the product of two elements of $SL_2(\Bbb Z/p\Bbb Z)$ is in $SL_2(\Bbb Z/p\Bbb Z)$ and for any element of $SL_2(\Bbb Z/p\Bbb Z)$ its inverse is in $SL_2(\Bbb Z/p\Bbb Z)$. Or use the shortcut and show for any $A,B\in SL_2(\Bbb Z/p\Bbb Z)$ that $A^{-1}B\in SL_2(\Bbb Z/p\Bbb Z)$ $\endgroup$ Commented Feb 1, 2016 at 14:04
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    $\begingroup$ Do you know the number of elements of $GL_2[\mathbb Z/p\mathbb Z]$? $SL_2$ is just the kernel of the determinant map, so counting its elements is fairly easy from there. $\endgroup$ Commented Feb 1, 2016 at 14:14
  • $\begingroup$ This could be useful (especially combined to Thomas's comment) : math.stackexchange.com/questions/1399406/… $\endgroup$
    – Arnaud D.
    Commented Feb 1, 2016 at 14:19

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Let $\mathbf{F}_p$ denote the field with $\mathbf{Z}/p$. Note that there is a homomorphism

$$ \textrm{det}: \textrm{GL}_n (\mathbf{F}_p) \to \mathbf{F}_p ^* $$ given by taking the determinant of the matrix. Note that the kernel of the homomorphism is exactly $\textrm{SL}_n (\mathbf{F}_p)$. Furthermore, this map is surjective. Therefore, we can use the First Isomorphism Theorem to count the number of elements in the special linear group.

$$|\textrm{SL}_n (\mathbf{F}_p) |= \frac{|\textrm{GL}_n (\mathbf{F}_p)|}{|\mathbf{F}_p ^*|} $$

The formula for the number of elements of $\textrm{GL}_n$ can easily be deduced by first counting it in the $n = 2$ case, which incidentally, is the case that you're interested in.

How can we create a $2 \times 2$ matrix that has determinant nonzero? Well, that's equivalent to being invertible, which is equivalent to the columns being linearly independent. The first column is free; anything nonzero will work, which gives $p^2-1$ choices, since there are $p$ choices for each of the two entries. The next column needs to be chosen outside of the span of the first columm, which has $p$ elements. Therefore, we have $p^2 - p$ choices for the second column.

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  • $\begingroup$ should it be $p^3 -p$ if we have $ p \times (p^2-1)$ $\endgroup$
    – Gino
    Commented Feb 1, 2016 at 14:44
  • $\begingroup$ The span of a single vector $v$ consists of $\alpha v$, where $\alpha$ comes from your field, which in this case is $\mathbf{F}_p$, which has $p$ elements. $\endgroup$
    – Future
    Commented Feb 1, 2016 at 17:37
  • $\begingroup$ @Thomas I have done this thank you. $\endgroup$
    – Future
    Commented Feb 1, 2016 at 17:38

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