If $G/Z$ is a product of $p$-groups... Let $G$ be a finite group. Suppose $Z \le Z(G)$ and $G/Z \cong A\times B$, where $A$ is a $\pi$-group and $B$ is a $\pi'$-group for some set of primes $\pi$. Is it true that $G\cong C \times D$, where $C$ is a $\pi$-group and $D$ is a $\pi'$-group for the same set of primes $\pi$?  
 A: Hall's version of Sylow's theorem handles this.
Background: A Hall π-subgroup is a π-subgroup whose index is a π′-number.  If every composition factor of G is either a π-subgroup or a π′-subgroup (that is, G is π-separable), then G posseses Hall π-subgroups.  If furthermore every composition factor is either a p-subgroup for some p in π or a π′-subgroup (that is G is π-solvable), then in fact all Hall π-subgroups of G are conjugate and all Hall π′-subgroups of G are conjugate.
Reduction: The condition is simply that both a Hall π- and π′-subgroup are normal.  Without loss of generality, Z = Z(G), since if a group has a normal Hall π-subgroup, then so does every quotient.
Proof: Write $G/Z = A/Z \times B/Z$ where $\gcd(|A/Z|,|B/Z|)=1$. A has a composition series refining A ≥ Z ≥ 1 with factors that are π and abelian.  Hence A is π′-solvable, and all of its Hall π-subgroups are conjugate.  Similarly B is π-solvable, and all of its Hall π′-subgroups are conjugate.
Let A0 be a Hall π-subgroup of A, and let B0 be a Hall π′-subgroup of B.  Every G-conjugate of A0 lies in A since A is normal, but then by Hall's theorem, such a G-conjugate is already A = A0Z-conjugate.  Hence $A_0^g = A_0^{az} = A_0$, and A0 is normal in G.  Similarly B0 is normal in G, and so G = A0 × B0. $\square$
A: To provide some context for this problem, I needed it as a lemma in order to do part b) of problem 5A.8 in Martin Isaac's book "Finite Group Theory". In page 151 of his book Isaacs gives the definition for the Schur multiplier and states (without proof of course) the existence and uniqueness of a largest second component in a defining pair and also that any other second component is a homomorphic image of that one. Assuming this, he then goes on and asks to prove  


*

*$|M(A\times B)| \geq |M(A)||M(B)|$

*if $(|A|,|B|)=1$, then $M(A \times B) \cong M(A) \times M(B)$


The first part is easy:
Suppose that $C$, $D$ are Schur representation groups for $A$ and $B$ respectively. Then 
$C \times D/M(A) \times M(B) \cong C/M(A) \times D/M(B) \cong A \times B$. 
Also, $(C \times D)’ = C’ \times D’$, while $Z(C \times D)=Z(C) \times Z(D)$. 
Hence $(C \times D)’ \cap Z(C \times D) \cong (C’ \cap Z(C)) \times (D' \cap Z(D))$, which contains $M(A) \times M(B)$ as a subgroup. Thus we see that $(C \times D,M(A) \times M(B))$ is another defining pair for $A \times B$. 
This shows that $M(A) \times M(B)$ is a homomorphic image of $M(A \times B)$. In particular, it shows that $|M(A) \times M(B)|$ divides $|M(A \times B)|$ and the desired inequality follows.
Under the assumption that $(|A|,|B|)=1$ the exercise can be used as a lemma to go all the way. If $E$ is a Schur representation group for $A \times B$, then there exist groups $C_0$, $D_0$ such that $E \cong C_0 \times D_0$. 
According to your proof, $C_0 \leq C$ and $D_0 \leq D$ so that $|E|$ divides $|C||D|$, which implies that $|M(A \times B)|$ divides $|M(A) \times M(B)|$. 
By the first part then, $|M(A \times B)|=|M(A) \times M(B)|$ and the homomorphism is forced to be an isomorphism.
There is of course some cheating behind all this, because nothing is said of the existence of a unique largest second component, such that all other second components are homomorphic images of that one. Nevertheless, I thought this was an interesting problem for someone to try.  
