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 (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?

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