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Count the number of $n\times n$ invertible matrices modulo $26$.

So far I am aware that a matrix is invertible if and only if its columns are linearly independent. I am also aware that the number of choices for the first column of an $n\times n$ matrix modulo $13$ is $13^n$, the number of choices for the 2nd column (given the first) is $(13^n)-13$ etc etc.

Similarly I also know the number of $n\times n$ invertible matrices modulo $2$.

So my question is this:

Using the Chinese remainder theorem, how can I use the fact that I know the number of nxn invertible matrices modulo $13$, and the number of invertible matrices modulo $2$ to find the number of invertible matrices modulo $26$?

Apologies for the length of this, and thank you in advance.

Jyrki Lahtonen
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    Pick an invertible $n\times n$ matrix $A$ modulo $2$, pick an invertible $n\times n$ matrix $B$ modulo $13$. The CRT says there is a unique matrix $C$ modulo $26$ such $C\equiv A\pmod 2$ and $C\equiv B\pmod{13}$. What can you say about the determinant of $C$ (modulo $26$ of course)? – Jyrki Lahtonen Feb 14 '16 at 18:21
  • well from this we know that C is also invertible modulo 26? –  Feb 14 '16 at 18:24
  • I'm not too sure what I can say about the determinant however, apart from the fact it is non zero –  Feb 14 '16 at 18:25
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    Correct. It goes the other way, too. If $C$ is invertible mod $26$, then by reducing it modulo two you get an invertible $A$, and and invertible $B$ by reducing it modulo $13$. – Jyrki Lahtonen Feb 14 '16 at 18:25
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    Oh, $\det C\equiv \det A\pmod 2$ and $\det C\equiv\det B\pmod{13}$, so $\det C$ is coprime to $26$ if and only if $\det A$ is odd and $\det B$ is not divisible by $13$. – Jyrki Lahtonen Feb 14 '16 at 18:27
  • Thank you very much Jyrki! –  Feb 14 '16 at 18:28

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