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

Question

It is well known that the automorphism group of the alternating group $A_6$ is $P\Gamma L_2(9)$. There are three different index $2$ subgroups of $P\Gamma L_2(9)$, namely the symmetric group $S_6$, the projective general linear group $PGL_2(9)$, and the Mathieu group $M_{10}$. By checking the ATLAS (http://brauer.maths.qmul.ac.uk/Atlas/v3/), the Schur multipliers of those groups $A_6,S_6, PGL_2(9)$ are the cyclic group $Z_6$. What about the Mathieu group $M_{10}$?

Also, I don't know why, in the following book, Page 302, Table 4.1, the Schur multiplier of the group ${\sf C}_2(2)$ is listed to be the cyclic group $Z_2$?

Gorenstein, Daniel, Finite simple groups. An introduction to their classification, Moskva: Mir. 352 p. R. 2.50 (1985). ZBL0672.20010.

Answer 1

$H_2(M_{10},\mathbb Z)\cong H^2(M_{10},\mathbb C^\times)\cong H^3(M_{10},\mathbb Z) = \oplus_{ p | 720} H^3(M_{10},\mathbb Z)_{(p)}$ with $p\in\lbrace 2,3,5\rbrace$. 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 $\mathbb Z_3^2\subset M_{10}$. This invariant subgroup is the full group $H^3(\mathbb Z_3^2)\cong\mathbb Z_3$ (see below), so $H_2(M_{10},\mathbb Z)\cong\mathbb Z_3$.

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 $M_9=\mathbb Z_3^2\rtimes Q_8$ 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 $H^\ast(\mathbb Z_3^2,\mathbb Z_3) = \mathbb Z_3[x,y]\otimes\Lambda(u,v)$ 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 $\mathbb Z\hookrightarrow\mathbb Z\twoheadrightarrow\mathbb Z_3$ is surjective and $Q_8$-equivariant, with image $\langle\delta(uv)\rangle$, 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), $H_2(M_9)\cong H^3(\mathbb Z_3^2)^{Q_8}\cong\mathbb Z_3$.