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Let $A$ and $B$ be subsets of a group $G$. The product $AB$ is called direct (and we denote it by $A \cdot B$, e.g., see this) if the representation of each element $x$ of $AB$ as $x=ab$, $a\in A$, $b\in B$ is unique (equivalently $A^{-1}A \cap BB^{-1}=\{1\}$, where $A^{-1}:=\{ a^{-1}:a\in A\}$).

Question. Let $G$ be a group and $A,B,C$ be subsets such that $G=A \cdot B=A \cdot C$. Then, is it true that $|B|=|C|$?

($|.|$ denotes the cardinal number)

Discussion. If $G$ is finite, then the answer is positive, since $|A \cdot B|=|A||B|=|A||C|$. If $1\in A$ (without loss of generality) and $A^{-1}A \subseteq AA^{-1}$, then we can define an injective (projection) map from $B$ to $C$ and also $C$ to $B$. Therefore, if there is a negative answer, it should be through infinite non-abelian groups $G$ (and infinite subsets $A$ with $A^{-1}A \nsubseteq AA^{-1}$).

Also see $AA^{-1} \subseteq A^{-1}A$ for every infinite subset $A$ of $G$ .

M.H.Hooshmand
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  • It is not true that $|AB|=|A||B|$, what it is realy true is that $|AB|=\frac{|A||B|}{|A\cap B|}$, see https://math.stackexchange.com/questions/2469369/showing-that-lvert-h-k-rvert-frac-h-k-h-cap-k?noredirect=1&lq=1 – Marcos Feb 24 '22 at 16:02
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    -@Marcus The intersection is a singleton (because the product is direct). – M.H.Hooshmand Feb 24 '22 at 16:25
  • right, in that case is true, but you have to say it in the question. Also, in this case $|A||B|=|A||C|$ and then $|B|=|C|$ – Marcos Feb 24 '22 at 16:53
  • The notation $A\cdot B$ means the product is direct. – M.H.Hooshmand Feb 24 '22 at 17:14
  • then? What is the problem? It is just a one line solution – Marcos Feb 24 '22 at 17:19
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    @Marcos: Not in the infinite case. If $|A|$ is infinite, then $|A||B|=\max{|A|,|B|}$. In principle, you could have an infinite $A$ and two finite $B$ and $C$ of distinct size with $|A||B|=|A||C|$. Likewise, you could have all three infinite, with $|A|$ of cardinality strictly larger than both $B$ and $C$, but with $|B|\neq|C|$. As to "saying it in the question", it says so in the question, prominently and at the top. The question also says the case of $G$ finite is trivial. Are you reading the question before declaring it trivial? – Arturo Magidin Feb 24 '22 at 17:53
  • Please see the discussion. We need an explicit example, do you have any? – M.H.Hooshmand Feb 24 '22 at 18:00
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    @M.H.Hooshmand Who are you addressing? Please use the @ if you are replying to a particular user. – Arturo Magidin Feb 24 '22 at 18:27
  • Sorry, ok, @Marcos. Did you see the discussion? – M.H.Hooshmand Feb 24 '22 at 18:30
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    That’s probably far too elementary an attempt, but there are natural maps $f: B \rightarrow C$ and $g: C \rightarrow B$ (where $b \in Af(b)$, $c \in Ag(b)$, so that $b \in AAg(f(b))$… which may not be enough. That kind of manipulations also show that if $A^{-1} \cdot B$ is direct, then $f$ is injective. (Thus if $A^{-1}\cdot B$ and $A^{-1} \cdot C$ are direct, then $f,g$ are injective thus $|B|=|C|$). – Aphelli Jul 15 '22 at 11:40
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    Isn't this the same question that you asked, and was answered, on MathOverflow a few years ago? – Jeremy Rickard Aug 04 '22 at 07:46
  • -@Jeremy Rickard. Oh yes, that's right. I forgot it. Thanks. – M.H.Hooshmand Aug 10 '22 at 14:47

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