Why is $SL(n,\mathbb{Z})$ a group under matrix multiplication and not a monoid like General Linear $GL(n,\mathbb{Z})$?
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First of all, $\mathrm{GL}_n(\mathbb{Z})$ is in fact a group, not just a monoid. It is the unit group of the ring $M_n(\mathbb{Z})$.
The reason that the inverse of a matrix in $\mathrm{GL}_n(\mathbb{Z})$ still has integer entries is the formula $$ A^{-1}=\frac{1}{\det(A)}\mathrm{adj}(A) $$ where $\mathrm{adj}(A)$ is the adjugate (or classical adjoint) of $A$. This matrix has integer entries, and if $A\in \mathrm{GL}_n(\mathbb{Z})$ then $\det(A)=\pm 1$, so $A^{-1}$ also has integer entries.
carmichael561
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I thought elements of $GL_n(\mathbb{Z})$ only had to have nonzero determinant, not necessarily a determinant equal to $\pm1$? – M10687 Aug 10 '16 at 21:03
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2No, the determinant must be $\pm1$. This is a common misconception. – carmichael561 Aug 10 '16 at 21:06
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Yeah, I just read about the definition over commutative rings. My mistake. – M10687 Aug 10 '16 at 21:06
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2@M10687 I had the misconception, too. See here. – Caleb Stanford Aug 10 '16 at 21:08
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Thanks for your reply. So what is the difference between GL(n,\mathbb{Z}) and SL(n,\mathbb{Z})? I thought the difference is the determinant value. – user128949 Aug 10 '16 at 21:30
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$\mathrm{SL}_n(\mathbb{Z})$ is the subgroup of $\mathrm{GL}_n(\mathbb{Z})$ consisting of matrices with determinant equal to $1$. So the difference is the determinant value. – carmichael561 Aug 10 '16 at 21:31
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And general linear group may have matrices with determinant -1? – user128949 Aug 10 '16 at 21:34
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Yes, exactly. For instance $\begin{bmatrix}1&0\0&-1\end{bmatrix}$ is in $\mathrm{GL}_2(\mathbb{Z})$ but not $\mathrm{SL}_2(\mathbb{Z})$. – carmichael561 Aug 10 '16 at 21:35
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Thanks so much. I am new to this topic. – user128949 Aug 10 '16 at 21:39