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$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Out{Out}\DeclareMathOperator{\cd}{cd}\DeclareMathOperator\SL{SL}$For a group $G$, the cohomological dimension of $G$ over the ring $R$, denoted by $\cd_R(G)$, is given by $$\cd_R(G)=\max\{n : H^n(G;M)\neq 0 \hspace{1mm} \mbox{for some} \hspace{1mm} RG\mbox{-module }M\}$$ However, $\Aut(F_n)$ is a discrete group of automorphisms of a free group with $n$ generators. The quotient by inner automorphisms is the outer automorphism group of a free group, denoted by $\Out(F_n)$. I would like to know the numbers $\cd_{\mathbb{Q}}\bigl(\Aut(F_n)\bigr)$, $\cd_{\mathbb{Q}}\bigl(\Out(F_n)\bigr)$, $\cd_{\mathbb{Q}}\bigl(\SL_n(\mathbb{Z})\bigr)$ and $\cd_{\mathbb{Z}}\bigl(\Aut(F_n)\bigr)$, $\cd_{\mathbb{Z}}\bigl(\Out(F_n)\bigr)$, $\cd_{\mathbb{Z}}\bigl(\SL_n(\mathbb{Z})\bigr)$.

P.S. I have googled these numbers, but I mostly found ‘virtual cohomological numbers’.

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    $\begingroup$ If a group has non-trivial torsion it has infinite cohomological dimension over Z and so virtual cohomological dimension is a better way to go. $\endgroup$
    – Benjamin Steinberg
    Commented Dec 28, 2022 at 12:48
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    $\begingroup$ The groups you list always have torsion when $n\ge2$. The virtual cohomological dimension is defined to be the cohomological dimension of a torsion-free subgroup of finite index (if there is one), or $\infty$ (if not). One always has $$\mathrm{cd}_{\mathbb Q}(G)\le \mathrm{vcd}_{\mathbb Z}(G)\le\mathrm{cd}_{\mathbb Z}(G).$$ These facts may help you extract the information you need from the results about vcd that you have already found. $\endgroup$
    – Peter Kropholler
    Commented Dec 28, 2022 at 15:08
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    $\begingroup$ But for $\mathrm{cd}_\mathbf{Q}$ one also needs, in these cases, to know that it is at least equal to $\mathrm{vcd}_\mathbf{Z}$ (this is probably false in general?) and probably the way the lower bound for $\mathrm{vcd}_\mathbf{Z}$ is obtained also provides a lower bound for $\mathrm{cd}_\mathbf{Q}$. I'm not sure about Aut/Out($F_n$), but for $\mathrm{SL}_n(\mathbf{Z})$ this is indeed $n(n-1)/2$ (lower bound being given by the inclusion of the upper unipotent group). $\endgroup$
    – YCor
    Commented Dec 28, 2022 at 16:00
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    $\begingroup$ It is not so easy to find groups of finite vcd for which $\mathrm{cd}_{\mathbb{Q}}$ is not equal to $\mathrm{vcd}_{\mathbb{Z}}$: I don't know that this is the case for the groups asked about, but it would be surprising if it wasn't. $\endgroup$
    – IJL
    Commented Dec 30, 2022 at 15:27
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    $\begingroup$ @archipelago Ah, yes, I misread what cd_Q was. Actually, I guess the result of Harer they state in the first paragraph, using the Steinberg module, shows cd_Q = vcd_Z for mapping class groups. $\endgroup$
    – Matt Zaremsky
    Commented Dec 31, 2022 at 12:11

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All the answers are contained in the comments by now, but let me compile everything together to be more official.

The easy part is that for all three groups, $\textrm{cd}_\mathbb{Z}$ is infinite, since the groups contain torsion.

As for $\textrm{cd}_\mathbb{Q}$, the answers are that $\textrm{cd}_\mathbb{Q}(\textrm{Aut}(F_n))=2n-2$, $\textrm{cd}_\mathbb{Q}(\textrm{Out}(F_n))=2n-3$, and $\textrm{cd}_\mathbb{Q}(\textrm{SL}_n(\mathbb{Z}))=n(n-1)/2$.

Here are a few more details: In each case, this value equals the virtual cohomological dimension of the group. The vcd is always an upper bound, so the point is that for all these groups it is also a lower bound, which can be shown by exhibiting some subgroup whose $\textrm{cd}_{\mathbb{Q}}$ is known to equal the vcd; call this a witness subgroup. For $\textrm{Aut}(F_n)$, it easy to see that it contains $\mathbb{Z}^{2n-2}$ as a witness (just consider all the automorphisms that fix $x_1$ and send each $x_i$ ($i\ne 1$) to $x_1^{m_i}x_i x_1^{n_i}$ for some $m_i,n_i\in\mathbb{Z}$). Similarly, for $\textrm{Out}(F_n)$ we have $\mathbb{Z}^{2n-3}$ as a witness. Finally, for $\textrm{SL}_n(\mathbb{Z})$, the subgroup of strictly upper triangular matrices is a witness.

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