# Subgroups written as products

Suppose a finite group $G$ is the product of two of its proper subgroups $G=AB$. Assume also that $A\lhd G$ and that $A,B$ have relatively prime orders. Isn't it true that any subgroup $H$ of $G$ can be written as $H=(H\cap A)(H\cap B)$?

-
You might want to look up Goursat's Lemma. It gives a characterisation of subgroups of the direct product of groups with respect to the Fibre Product, I believe. – user1729 Jul 23 '12 at 15:24
What does the notation $A\lhd G$ mean? – celtschk Jul 23 '12 at 17:52
@celtschk: It means "$A$ is a normal subgroup of $G$". That is, $A$ is a subgroup of $G$ such that $gag^{-1}\in A$ for all $a\in A$ and $g\in G$. – user1729 Jul 23 '12 at 18:54
@user1729: Thank you. While I knew the concept of normal subgroups, I didn't know that there's a symbol for it. Is there somewhere a complete list of common notations, per field, where such things could be looked up? – celtschk Jul 23 '12 at 19:43
@celtschk: Probably, but I don't know of one. However, if you pick up your favourite introductory text to some area of mathematics then it will most likely have a bit at the back where all the symbols used are defined. – user1729 Jul 24 '12 at 17:24

While Qiaochu's answer settles the original question, it might be worth noting that every subgroup $H$ of $AB$ under these hypotheses can be written in the form $(H \cap A)(H \cap B^{x})$ for some $x \in A.$ Let $\pi$ be the set of prime divisors of $A$. Then $H/(H \cap A)$ is isomorphic to a subgroup of $B$, so is a $\pi^{\prime}$-group. By the Schur-Zassenhaus theorem, we have $H = (H \cap A)C$ for some subgroup $C$ of $H$ with $(H \cap A) \cap C = 1.$ By Schur-Zassenhaus again, we have $C^{g} \leq B$ for some $g \in G.$ Write $g = ab$ for some $a \in A, b \in B.$ Then $C^{a} \leq B.$ Hence $H^{a} = (H \cap A)^{a}C^{a} \leq (H^{a} \cap A)(H^{a} \cap B) \leq H^{a}.$ Setting $x = a^{-1}$, conjugating by $x$ gives $H = (H \cap A)(H \cap B^{x}).$

-
There is a solubility condition in Schur-Zassenhaus I believe. – the_fox Jul 24 '12 at 1:46
Well, the Schur Zassenhaus Theorem require that $A$ or $G/A$ be solvable when $A$ is a normal subgroup whose order and index in $G$ are coprime. But since $|A|$ an $|G/A|$ are coprime, at least one of them has odd order, so is solvable ( though this does ultimately rely on the Feit-Thompson odd order theorem) – Geoff Robinson Jul 24 '12 at 11:45

No. For example, $H$ may be a nontrivial conjugate of $B$. If $B$ is not normal such a conjugate always exists and by hypothesis $H \cap A$ will be trivial and $H \cap B$ will be strictly smaller than $H$.

For an explicit example, take $G = S_3, A = A_3$, let $B$ be the subgroup generated by a transposition, and let $H$ be the subgroup generated by another transposition. Then $(H \cap A)$ and $(H \cap B)$ are both trivial.

-
Thanks. If $G$ has that property though, i.e. all proper subgroups can be factorised in this way, can anything be said about the structure of $G$? – the_fox Jul 23 '12 at 15:22
Silly question. If G has that property then $G=A\times B$. – the_fox Jul 23 '12 at 15:28
Can you give a non-trivial example of a group with that property, @Stefanos? I don't think it is that easy, if there exists at all... – DonAntonio Jul 23 '12 at 15:40
@Don: take $C_p \times C_q$ for distinct primes $p, q$. – Qiaochu Yuan Jul 23 '12 at 15:52
Thanx. I was dealing with something related to infinite groups and completely forgot the good'ol finite cyclic ones...:) – DonAntonio Jul 23 '12 at 16:05