Let $G$ be a finite group such that if $A, B\le G$ then $AB\le G$. Prove $G$ is a solvable.

Question

Let $G$ be a finite group satisfying the following property: (*) If $A, B$ are subgroups of $G$ then $AB$ is a subgroup of $G$. Prove $G$ is solvable.

So I felt like a good place to start is to let $|G| = p_1^{a_1}p_2^{a^2}...p_r^{a_r}$ such that each $p_i$ is a distinct prime, and then Let for each $i = 1, 2, ..., r$ Let $H_i$ be a Sylow$_{p_i}$ subgroup. So $G = H_1H_2..H_r$ and each $H_i$ is solvable since all $p$-groups are solvable.

I was then thinking that a good idea since I know by Burnside's Theorem that $H_iH_j$ is solvable, to try and use induction and prove that if $H_1H_2...H_m$ is solvable then $H_1H_2...H_mH_{m+ 1}$ is sovable, but I can't figure it out.

Any help will be greatly appreciated.

Answer 1

Here's another proof, now that you have the answer. Let $x$ and $y$ be elements of coprime order. Then $\langle x\rangle\langle y\rangle$ is a subgroup, and has order $o(x)\cdot o(y)$. Thus any two elements of coprime order commute.

It turns out this is enough to prove the result: a finite group $G$ is nilpotent if and only if every $2$-generated subgroup is nilpotent.

But let's not use that, and let $z$ be an element in the centre of some Sylow $p$-subgroup of $G$. Consider $C_G(z)$. It contains all elements of order prime to $p$, and also contains a Sylow $p$-subgroup of $G$, thus is all of $G$. Thus $z\in Z(G)\neq 1$.

Since your condition is inherited by quotients, we see that $G$ must possess a central series, hence is nilpotent.

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Edit: some more on the structure of these groups. In general (i.e., infinite groups), they are locally nilpotent. The finite ones were mostly classified by Iwasawa in 1941, but there were some gaps that were filled by Napolitani and Janko.

A $p$-group is quasihamiltonian (i.e., $AB$ is a subgroup for all $A,B\leq G$) if and only if

1. It has an abelian normal subgroup $N$ such that $G/N$ is cyclic, and there exists a generator $x$ of $G/N$, and an integer $n$, such that $a^x=a^{1+p^n}$ for all $a\in A$, or
2. $p=2$, and $G$ is the direct product of $Q_8$ and an elementary abelian $2$-group.

A subgroup $H$ such that $HK=KH$ (i.e., $HK$ is a subgroup) for all $K\leq G$ is called a permutable subgroup. The above groups are those where all subgroups are permutable. There's been recent research on groups where lots of, but not all, subgroups are permutable.

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Edit 2: here's another proof, which given you mention Burnside's $p^\alpha q^\beta$ theorem, might have been the intention of your tutor (assuming you are doing a course).

Let $x\in G$ have order $p$ for some $p\mid |G|$. By the above method, we see that $y\in C_G(x)$ for all elements $y\in G$ of order prime to $p$. But now this means that $|G:C_G(x)|$ is a power of $p$. By that useful theorem you did while proving Burnside's $p^\alpha q^\beta$ theorem, $G$ cannot be a non-abelian simple group. Thus $G$ possesses a normal subgroup $N$. The condition is inherited by both $N$ and $G/N$, so by induction both of these are soluble. Thus $G$ is soluble.

Answer 2

Claim: All Sylow subgroups of $G$ are normal.

Proof: Suppose $P_1,P_2$ are different Sylow $p$-subgroups of $G$. Then the group $P_1P_2$ is a $p$-group that is strictly larger than $P_1$, contradiction.

Corollary: $G$ is solvable.