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 $H=(H\cap A)(H\cap B)$?
Subgroups written as products
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0You 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. – 2012-07-23
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0What does the notation $A\lhd G$ mean? – 2012-07-23
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0@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 $g\in G$. – 2012-07-23
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0@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? – 2012-07-23
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0@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. – 2012-07-24
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0@user1729: That method has the problem that it describes the notation used in *that book* which may be (a) just one of several common notations (think of $\mathbb{N}$ vs. $\mathbb{N_0}$ or $\subset$, $\subseteq$, $\subsetneq$), (b) a notation where no common notation exists, (c) (less likely in a beginner's text book, but not impossible) a notation which differs from the common notation, but which the author considers superior (although in that case I'd hope the authour would at least tell that the notation is non-standard). Generally you'll have no indication whether (a) or (b) applies. – 2012-07-24
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0Sure, but that will always be the case. Alternatively, ask you local, friendly, neighbourhood professor (or on here!) if you don't know or aren't sure. – 2012-07-24
2 Answers
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 $H = (H \cap A)(H \cap B^{x}).$
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0There is a solubility condition in Schur-Zassenhaus I believe. – 2012-07-24
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0Well, 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) – 2012-07-24
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.
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0Thanks. 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 $G$? – 2012-07-23
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0Silly question. If G has that property then $G=A\times B$. – 2012-07-23
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0Can 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... – 2012-07-23
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1@Don: take $C_p \times C_q$ for distinct primes $p, q$. – 2012-07-23
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0Thanx. I was dealing with something related to infinite groups and completely forgot the good'ol finite cyclic ones...:) – 2012-07-23
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0@Don: There are many such groups which are not Abelian. For example $S_{3}$ and $A_{4}$ are small examples – 2012-07-23
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0@GeoffRobinson, sorry: I don't understand your comment. I was trying to come up with direct products of groups all the subgroups of which are again direct products of subgroups of the factors. What has abelian/nonabelian and $\,S_3\,,\,A_4\,$ to do with this? – 2012-07-23
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0@Don: I misundrstood. I thought you were asking for groups $G = AB$ such that $A \lhd G$ and ${\rm gcd}(|A|,|B|) = 1$, of which $S_{3}$ and $A_{4}$ are examples. – 2012-07-23