From "Construction of Finite Groups" article :
"To construct minimal Frattini extensions of $H$ we have to consider all suitable irreducible $H$-modules $M$. Clearly, $M$ can be viewed as an irreducible $F_p$ $H$-module for some prime $p$."
Why is $M$ considered as $F_p$ $H$-module ? it is not clear for me.
For a finite group $H$, irreducible $\mathbb{Z}[H]$-modules $V$ are vector spaces over some finite field $\mathbb{Z}/p\mathbb{Z}$ since for each prime $q$, $$qV = \{ \underbrace{v + v + \ldots + v }_{q \text{ summands}} : v \in V \}$$ is a submodule, so is either $qV =V$ or (as is the case for $q=p$) $qV=0$. In case one does not already assume $V$ is finite, one should also keep in mind that for any $v$, the $H$-submodule $\langle v \rangle$ is a finitely generated abelian group, and so cannot be divisible (one cannot have $q\langle v \rangle = \langle v \rangle$ for every prime $q$).
At any rate, if $pV=0$ then $p$ acts as $0$ on $V$, so we can replace integer coefficients with integer mod $p$ coefficients, that is $V$ is naturally $\mathbb{Z}/p\mathbb{Z}[H]=\mathbb{F}_pH$ module.
From a different point of a view: $M$ is a nilpotent (being contained in the Frattini) minimal normal subgroup of a finite group, so it is a finite direct product of copies of $C_p$ for some prime $p$.