What is the Schur multiplier of the Mathieu group $M_{10}$

$H_2(M_{10},mathbb Z)cong H^2(M_{10},mathbb C^times)cong H^3(M_{10},mathbb Z) = oplus_{ p | 720} H^3(M_{10},mathbb Z)_{(p)}$ with $pinlbrace 2,3,5rbrace$. A $p$-primary component is isomorphic to the set of $M_{10}$-invariant elements of $H^3(text{Syl}_p(M_{10}))$. We can check that $M_{10}$ has semi-dihedral Sylow 2-subgroups and cyclic Sylow 5-subgroups (Wikipedia), both of whose Schur multiplier is trivial. So we're left with the 3-primary component. We can check that $M_{10}$ has elementary abelian Sylow 3-subgroups (Wikipedia), all isomorphic to $mathbb Z_3^2$, and thus $H^3(M_{10})_{(3)}cong H^3(mathbb Z_3^2)^{H/mathbb Z^2_3}$ by Swan's theorem, where $H$ is the normalizer of $mathbb Z_3^2subset M_{10}$. This invariant subgroup is the full group $H^3(mathbb Z_3^2)congmathbb Z_3$ (see below), so $H_2(M_{10},mathbb Z)congmathbb Z_3$.

To see that $H^3(mathbb Z_3^2)^{H/mathbb Z_3^2}cong H^3(mathbb Z_3^2)$, we first note that our $H$ is a maximal subgroup of order 72 so we may as well take it to be the Mathieu group $M_9=mathbb Z_3^2rtimes Q_8$ sitting naturally inside $M_{10}$, where $Q_8$ acts as the faithful two-dimensional irreducible representation over $mathbb Z_3$. So we need to check that $H^3(mathbb Z_3^2)^{Q_8}cong H^3(mathbb Z_3^2)$. For the standard generators ${I,J}$ of $Q_8$ and basis ${a,b}$ of $mathbb Z_3^2$ we have $I(a)=a+b$, $I(b)= a-b$, $J(a)=-a+b$, and $J(b)=a+b$. Noting that $H^ast(mathbb Z_3^2,mathbb Z_3) = mathbb Z_3[x,y]otimesLambda(u,v)$ with $|x|=|y|=2$ and $|u|=|v|=1$, the element $uv$ is $Q_8$-invariant. The induced map $delta:H^2(mathbb Z_3^2,mathbb Z_3)to H^3(mathbb Z_3^2,mathbb Z)$ under the short exact sequence $mathbb Zhookrightarrowmathbb Ztwoheadrightarrowmathbb Z_3$ is surjective and $Q_8$-equivariant, with image $langledelta(uv)rangle$, and so $H^3(mathbb Z_3^2)^{Q_8}cong H^3(mathbb Z_3^2)$.

We've essentially shown that this agrees with the Schur multiplier of $M_9$ by the same technique (note that the Schur multiplier of $Q_8$ is trivial), $H_2(M_9)cong H^3(mathbb Z_3^2)^{Q_8}congmathbb Z_3$.