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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    $\begingroup$ I think it's called "factorization" for advertising purposes: I don't think even partly reduces the understanding of the group to simpler subgroups (and hence I wouldn't call "prime" such a marginal notion). Also note that a group without factorization $AB$ might have a (nontrivial) factorization $ABC$. (Actually, for every finite group $G$ that is not cyclic of prime power order, there exists $n$ and proper subgroups $G_1,\dots,G_n$ such that $G=G_1G_2\dots G_n$.) $\endgroup$
    – YCor
    Jul 29, 2019 at 19:26
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    $\begingroup$ Z/p^nZ is also "non factorizable" $\endgroup$ Jul 30, 2019 at 0:02
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    $\begingroup$ Finite groups $G$ with $|G|=p^n$ (up to cyclic groups) and $|G|=p^nq^m$ have a factorization ($p,q$ distinct primes). So for groups without factorization you have to look at groups those order has at least three distinct prime divisors. $\endgroup$
    – tj_
    Jul 30, 2019 at 5:23
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    $\begingroup$ For (finite) solvable groups you can always use Hall subgroups to find a factorization. $\endgroup$
    – j.p.
    Jul 30, 2019 at 5:58
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    $\begingroup$ Tarski Monsters give examples of infinite groups without factorizations. en.wikipedia.org/wiki/Tarski_monster_group $\endgroup$
    – Ian Agol
    Jul 30, 2019 at 21:02

2 Answers 2


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 $\Phi(G)$).

  • $\begingroup$ After your answer, we are reduced "in some sense" to classify the finite simple groups without factorization. Is there a non-abelian and non-sporadic finite simple group without factorization? If so, what are there? $\endgroup$ Jul 30, 2019 at 22:14
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    $\begingroup$ @SebastienPalcoux: Have you tried looking for an answer in the AMS memoir by Liebeck, Praeger, and Saxl? As Geoff notes in his answer, clearly it suffices to check when there is a factorization into a product of two maximal subgroups. And Liebeck-Praeger-Saxl classify these factorizations. $\endgroup$
    – spin
    Jul 30, 2019 at 22:41
  • $\begingroup$ As @spin says, the simple groups which admit no factorisation are classified by Liebeck-Praeger-Saxl. There is also the question about which of these admit a non-split extension by an irreducible (finite, possibly trivial) irreducible module $V$. $\endgroup$ Jul 31, 2019 at 8:18
  • $\begingroup$ @GeoffRobinson: I am not sure how to read the tables in this book. For example, $M_{22}$ appears in Table 6 (page 15) whereas it does not appear in the paper of Michael Giudici cited above. $\endgroup$ Jul 31, 2019 at 8:49
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    $\begingroup$ @SebastienPalcoux: The notation is explained in the book. Let $L = M_{22}$. The entry of $L$ in Table 6 is marked with (*) and it is stated that $G = L.2 = \operatorname{Aut}(L)$. We have $G = AB$ but $L \neq (L \cap A)(L \cap B)$ in this particular case, which is proven in 6.3 in the book $\endgroup$
    – spin
    Jul 31, 2019 at 9:08

The smallest non-abelian group without factorization is simple of order $1092$: it is $A_1(13)$.

Using the answer of Geoff and by browsing the book The maximal factorizations of the finite simple groups and their automorphism groups (by Martin W. Liebeck, Cheryl E. Praeger and Jan Saxl), the finite simple groups without factorization are the following:

  • Every cyclic group of prime order.
  • Alternating: none
  • Chevalley groups: $A_n(q)$ with $n=1$, $q \equiv 1 (\textrm{mod}\ 4)$ and $q\neq 5,9,29$, or with $n \ge 2$ even and $(n+1) | (q-1)$; $E_6(q)$; $E_7(q)$; $E_8(q)$; $F_4(q)$ with $q \neq 2^n$; $G_2(q) $ with $q>4 $ and $q \neq 3^n$.
  • Twisted Chevalley groups: $^2A_{2n}(q^2)$ except $(n,q) = (1,3),(1,5),(4,2)$; $^2B_2(2^{2n+1})$; $^2D_{2n}(q^2)$; $^3D_4(q^3)$; $^2E_6(q^2)$; $^2F_4(2^{2n+1})$; $^2G_2(3^{2n+1})$.
  • Sporadic: $J_1$, $M_{22}$, $J_3$, $McL$, $O'N$, $Co_3$, $Co_2$, $F_5$, $Ly$, $F_3$, $Fi_{23}$, $J_4$, $F_{3+}$, $F_2$, $F_1$.

Before any use: this list requires a double checking.


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