# Factor subset of finite group

Let $G$ be a group of order $n$ and $d$ a positive divisor of $n$. Is it true that there exists a subset $A$ of $G$ with $d$ elements and a subset $B$ such that $G=AB$ and $|AB|=|A||B|$ (equivalently, the product $AB$ is direct)?

Note: If $G$ has the property that there exists a subgroup of $G$ with order $d$ or $n/d$, then we can show that the answer is positive (see A question about finite groups (a weak version of the converse of Lagrange theorem)).

Now, is the answer positive for ${\rm PSL}(2,8)$, ${\rm PSL}(2,11)$, ${\rm PSL}(2,13)$, ${\rm SL}(2,11)$, ${\rm SL}(2,13)$, ${\rm PSL}(2,17)$, ${\rm A}_7$, ${\rm PSL}(2,19)$, ${\rm A}_5 \times {\rm A}_5$?

About $PSL_2(8)$ (in the last following comment, by Russ Woodroofe). If it has subgroups of indexes $6$, $14$ and $21$ with the property $P(*,*)$, then we can say that $PSL_2(8)$ has $P(*,*)$. Note that $PSL_2(8)$ has subgroups of indexes $6$ (that is solvable and so $P(*,*)$ is true for it), $14$ (that is also solvable) and $21$ (with order $84$). Now, is $P(*,*)$ valid for the last subgroup? (if yes, then $PSL_2(8)$ has $P(*,*)$).

• Even in the case where A and B are subgroups and AB = G, and |G| = |A||B|, one does not generally refer to this as a "direct product" unless both A and B are normal subgroups. – Marty Isaacs Jan 28 '14 at 18:50
• One can ask the question for quasigroups G and look at partitioning transversals. If there is a counterexample, I would imagine it is pretty small, and that one for groups would not be much larger. Gerhard "Ask Me About Binary Operations" Paseman, 2014.01.28 – Gerhard Paseman Jan 28 '14 at 19:01
• A mildly useful keyword might be Zappa-Szép product -- this is the case where $A$ and $B$ are both taken to be subgroups. See Wikipedia: en.wikipedia.org/wiki/Zappa–Szép_product (MathOverflow breaks the link at the -, so copy and paste into your browser unless you're interested in Frank, rather than Guido, Zappa.) – Russ Woodroofe Jan 28 '14 at 22:47

This is true for solvable groups, at least. We are given a group $G$ of order $ab$, where $a$ and $b$ are integers, and we want subsets $A$ and $B$ of $G$ such that $|A| = a$, $|B| = b$ and $G = AB$. We work by induction on $|G|$ to prove that $A$ and $B$ exist if $G$ is solvable. We can certainly assume that $|G| > 1$. Since $G$ is solvable, it has a subgroup $H$ of prime index $p$, and we can assume without loss that $p$ divides $b$. Then $|H| = (a)(b/p)$, so by the inductive hypothesis, we can write $H = AX$, where $|A| = a$ and $|X| = b/p$. Now let $T$ be a set of representatives for the right cosets of $H$ in $G$. Then $|T| = p$ and $G = HT = AXT$. Let $B = XT$, so $AB = G$. Also, $|B| = |XT| \le |X||T| = (b/p)p = b$. Equality must hold here since $AB = G$.

• Unless I'm mistaken, the same argument proves the statement for any group having a series of subgroups, each of prime index in the next. That takes care of e.g. $A_5$, and indeed any group with all composition factors either abelian or isomorphic to $A_5$. – Russ Woodroofe Jan 28 '14 at 23:02
• @RussWoodroofe Yes, that's right. – Marty Isaacs Jan 29 '14 at 16:55

You can handle several of the groups on the list from your edit with a modification of the argument of Marty Isaacs. For the rest, you can find guidance as to what a counterexample would look like.

Say that $G$ satisfies $P(a,b)$ if $ab = |G|$ and there are subsets $A$ and $B$ of cardinalities $a$ and $b$ such that $G = AB$. Write $P(*,*)$ when $G$ satisfies $P(a,b)$ for any $a,b$ with $ab = |G|$.

The following has exactly the same proof as the solvable argument.

Lemma: If $H$ is a subgroup of $G$ satisfying $P(a,b)$, and $[G:H] = t$, then $G$ satisfies $P(a,bt)$.

Then if $G$ has a maximal subgroup of prime index satisfying $P(*,*)$, then $G$ also satisfies $P(*,*)$. That handles $A_5$, $PSL_2(7)$, and $A_5 \times A_5$.

Moreover, if $G$ has a maximal subgroup of index $t$, then $G$ satisfies $P(a,b)$ for every pair $a,b$ with t dividing $b$ (and of course $ab=|G|$).

1. This lets us handle $A_6$, which has maximal subgroups of index 6, 10, and 15. If $ab = 60$ then either $a$ or $b$ is divisible by one of these.
2. $A_7$ then follows, since $A_6$ is a maximal subgroup of $A_7$ having (prime) index 7.
3. Similar (but slightly more involved) arguments to those for $A_6$ show that $A_8$ satisfies $P(*,*)$.
4. The first open case is $PSL_2(8)$, which has order $504 = 2^3 \cdot 3^2 \cdot 7$, and maximal subgroups with indices 9, 28, and 36. Since every subgroup satisfies $P(*,*)$, the possible places where $PSL_2(8)$ could fail $P(*,*)$ are $P(12, 42)$ and $P(21, 24)$.

By the way, GAP will calculate the relevant data for these arguments for you pretty easily. The following commands will do it (replace $G$ with your favorite group):
G:=PSL(2,7);; Print(Size(G), " = ", FactorsInt(Size(G)), "\n"); Set(List(MaximalSubgroups(G), x->Index(G, x)));

• That's good idea. It seems that the term "maximal" in the above propositions is not necessary, isn't it? – M.H.Hooshmand Jan 31 '14 at 19:40
• No, I don't think maximal is necessary. Indeed, you might get a bit of extra power by looking also at meet-irreducible (but not necessarily maximal) subgroups. – Russ Woodroofe Jan 31 '14 at 20:54
• Presumably the inductive approach that Marty started and you continue is likely to hit problems when we have an integer $d$ such that no proper subgroup of $G$ has order divisible by $d$ or $\frac{n}{d}.$ – Geoff Robinson Jul 5 '14 at 22:06
• @Geoff Robinson: Yes, sounds like a good place to look for a possible counterexample. – Russ Woodroofe Jul 7 '14 at 11:23
• As one can see in mathoverflow.net/questions/177747/…, there are such subsets for $|Psl(2,8)|=21\times 24$. So, $PSL(2,8)$ has $P(∗,∗)$. For other candidates see the link. – M.H.Hooshmand Aug 7 '14 at 19:33