A group G is said to have a *factorization* if there exist proper subgroups $A$ and $B$ such that $G = AB = \{ ab \ | \ a \in A, b \in B \}$. 


The paper [Factorisations of sporadic simple groups](https://doi.org/10.1016/j.jalgebra.2006.04.019) (by Michael Giudici) provides the classification of all the factorizations of the sporadic simple groups. We observe there that only $11$ sporadic simple groups (among $26$) admit a factorization. So the remaining $15$ (which are $J\_1$, $M\_{22}$, $J\_3$, $McL$, $O'N$, $Co\_3$, $Co\_2$, $F\_5$, $Ly$, $F\_3$, $Fi\_{23}$, $J\_4$, $F\_{3+}$, $F\_2$, $F\_1$) admit no factorization. 


Obviously, a cyclic group admits no factorization if and only if it is of prime power order.


**Question**: What are all the (other) finite groups without factorization?  

(Is there an *official* name for such a group?) 


*Proposition*: Let $G$ be a group without factorization. If the intersection of all the maximal subgroups of $G$ is the trivial subgroup then $G$ is simple.  

*proof*: Assume $G$ non-simple and let $N$ be a non-trivial proper normal subgroup of $G$. If every maximal subgroup of $G$ contains $N$ then their intersection also, which contradicts the assumption. So there is a maximal subgroup $M$ of $G$ not containing $N$. It follows that $NM=G$, contradiction. $\square$


*Sub-question*: must a non-simple finite group without factorization be cyclic (of prime power order)?


*Lemma*: A finite group $G$ admits a unique maximal subgroup $M$ iff it is cyclic of prime power order.  

*proof*: Let $g \in G \setminus M$ and $H = \langle g \rangle$. If $H \neq G$ then there must exist a maximal subgroup $M'$ of $G$ containing $H$, but $M=M'$ by assumption, so $g \in M$, contradiction. So $G = H$ is cyclic. It is moreover of prime power order by Chinese Remainder Theorem. $\square$


*Bonus question*: What do we know about the infinite groups without factorization?



Groups without factorization
This must be "well-known": If we have $G = AB$ when $G$ is a finite group, and $A,B$ are proper subgroups of $G$, then we may suppose that $A$ and $B$ are both maximal.


For if $A$ is not maximal, and $A < C$ with $C$ maximal, then we still have $G = CB,$ and $B \not \leq C$, so we may replace $A$ by $C$ and assume that $A$ is maximal.


Now if $B$ is not maximal, but $B < D$ with $D$ maximal, then we have $G = AD$ and $A,D$ are both proper and maximal, so we can replace $B$ by $D$ and assume, $A,B$ are both maximal.


Hence the finite group $G$ has no factorization if and only if $|AB| < |G|$ whenever $A$ and $B$ are maximal subgroups of $G$, ie if and only if $[G:A] > [B:A \cap B]$ whenever $A,B$ are maximal subgroups of $G$.


In particular, since the Frattini subgroup $\Phi(G)$ is the intersection of the maximal subgroups of $G$ we conclude that $G$ has a factorization if and only if $G/\Phi(G)$ has a factorization. 


Since $G/\Phi(G)$ has trivial Frattini subgroup, we conclude from the argument you gave in the question, that if $G$ has no factorization, then $G/\Phi(G)$ is a simple group. 


Later edit: By the way, this answers the subquestion in the negative. If we take a nonsplit extension $H$ of a non-Abelian finite simple group $G$ which itself has no factorization by a finite irreducible $G$-module $M$, then we obtain a non-simple finite group $H$ which is neither simple nor cyclic of prime-power order, yet there is is no factorization of $H$ (and there are examples of such non-split extensions- note that $\Phi(H) = M$ in that case. For example, $J\_{3}$ admits no factorization, and it has a (non-split) triple cover which also admits no factorization).


Even later edit: In fact, we can now further restrict all finite groups $G$ which admit no factorization. By a slight extension of the argument above, we deduce that $G$ has no factorization if and only if $G/X$ has no factorization whenever $X \lhd G$ with $X \leq \Phi(G)$. 


If G has no factorization, we know that $G/\Phi(G)$ is a simple group without a factorization. 


If $G/\Phi(G)$ is cyclic of prime order, then $G$ is cyclic, and $G$ must be a cyclic $p$-group for some prime $p$.


If $G/\Phi(G)$ is non-Abelian simple, then we may take $X \lhd G$ with 
$X < \Phi(G)$ and $[\Phi(G):X]$ minimal. Let $V = \Phi(G)/X.$


Then $V = \Phi(G/X)$ and $G/X$ is a non-split extension of the simple group $G/\Phi(G)$ by the irreducible $G/\Phi(G)$-module $V.$ 


Hence the finite group $G$ admits no factorization if and only if either $G$ is cyclic of prime power order or else $G/\Phi(G)$ is a non-Abelian simple group with no factorization which admits a non-split extension by a finite 
(possibly trivial) irreducible $G/\Phi(G)$-module $V$ occurring as a "top" $G$-chief factor within $\Phi(G)$ (that is to say $V$ occurs as $\Phi(G)/X$ where $X$ is maximal subject to being normal in $G$ and properly contained in $\Phi(G)$). 


