Q:

Number of invertible matrices modulo 26

The number of invertible matrices modulo $26$ can be computed by the Chinese Remainder Theorem. i.e. a matrix is invertible modulo 26 if it is invertible modulo $13$ and modulo $2$ which are given by the order of the groups $GL(n, \mathbb{Z}_{2})$ and $GL(n,\mathbb{Z}_{13})$: $$2^{n^2}(1-1/2)(1-1/2^2)\ldots(1-1/2^n)$$$$13^{n^2}(1-1/13)(1-1/13^2)\ldots(1-1/13^n)$$ and so the number of invertible $n\times n$ matrices is their product, i.e. $$26^{n^2}(1-1/2)(1-1/2^2)\ldots(1-1/2^n)(1-1/13)(1-1/13^2)\ldots(1-1/13^n)$$

My questions are, how does this use Chinese Remainder Theorem and how do the product of these quantities give us the number of matrices? What if a matrix is invertible modulo $13$ but not modulo $2$? Is it still invertible modulo $26$?

A:

The Chinese Remainder Theorem tells us that $\mathbb{Z}_{26}\cong\mathbb{Z}_2\times\mathbb{Z}_{13}$, which in turn is what tells us that $GL(n,\mathbb{Z}_{26})\cong GL(n,\mathbb{Z}_2\times\mathbb{Z}_{13})\cong GL(n,\mathbb{Z}_2)\times GL(n,\mathbb{Z}_{13})$. Then the result follows by the proof you have in your answer.
It is definitely not the case that a matrix which is invertible modulo $13$ but not invertible modulo $2$ will be invertible modulo $26$. For example, take the matrix with all $2$'s on its diagonal and zeros everywhere else. Then this matrix is in $GL(n,\mathbb{Z}_{13})$, is not in $GL(n,\mathbb{Z}_2)$, and is not in $GL(n,\mathbb{Z}_{26})$, as $2$ does not have a multiplicative inverse in $\mathbb{Z}_{26}$.

