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Let $A, B$ be two subgroups of a group $G$. Define $AB = \{ab\mid a \in A, b \in B\}$ and $BA = \{ba\mid b \in B, a \in A\}$. Prove that $AB\subset BA$ if and only if $BA\subset AB$.

I try to prove the statement. For $AB \subset BA$, I try to prove that $BA\subset AB$. Consider $x\in BA$, then there exist $b_1\in B$ and $a_1\in A$ so that $x=b_1a_1$. But I am stuck on how to show that $x\in AB$?

Hermi
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  • https://math.stackexchange.com/questions/1704893/when-hk-is-a-subgroup-why-kh-subseteq-hk is somewhat related, but not quite the same question. But maybe it can give you a nudge in the right direction? Same can be said of https://math.stackexchange.com/questions/1197732/prove-that-hk-is-a-subgroup-iff-hk-kh. And just to top it off, https://math.stackexchange.com/questions/1550446/how-to-prove-a-statement-with-two-if-and-only-if has your problem as a subproblem, but doesn't actually ask about it directly. – Arthur Sep 16 '23 at 21:33
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    Hint: $X\subset Y\iff X^{-1}\subset Y^{-1}.$ – Anne Bauval Sep 16 '23 at 21:35
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    Exactly! And I find my hint quite obvious. (At least $\implies$ is obvious by definition of $X^{-1},$ and is sufficient for our purposes. $\impliedby$ can be deduced from it, using $(X^{-1})^{-1}=X,$ but is useless here.) – Anne Bauval Sep 16 '23 at 21:45
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    @Hermi $X$ can be any subset of $G$, the point is $x\mapsto x^{-1}$ is a bijection (though you really just need it to be a function for $\implies$). – M W Sep 17 '23 at 00:29
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    @Hermi There's nothing strictly false in there that I see, but the whole part about showing $X\subseteq Y$ iff $X^{-1}\subseteq Y^{-1}$ is completely unnecessary, as Anne Bauval pointed out. The fact that inversion is a well-defined function immediately gives you that $X\subseteq Y$ implies $X^{-1}\subseteq Y^{-1}$ (you can really use this without any need to say anything in my opinion), and that is the only thing you use in your proof. – M W Sep 17 '23 at 05:25
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    You use $$x\in X^{-1}\implies x^{-1}\in X$$ and $$x=(x^{-1})^{-1},$$ which is a luxury. For any function $f:E\to F$ and any $X,Y\subset E,$ it is clear from the definition of $f(X)$ that $$X\subset Y\implies f(X)\subset f(Y).$$ – Anne Bauval Sep 17 '23 at 05:42
  • @AnneBauval Thank you! But I am not sure if we can get ` $(AB)^{-1}\subset (BA)^{-1}$ implies $B^{-1}A^{-1} \subset A^{-1}B^{-1}$? I am confused if we can use $(AB)^{-1}=B^{-1}A^{-1}$? We do not know $AB$ is a subgroup. – Hermi Sep 17 '23 at 17:09
  • @AnneBauval I improved my proof. Can you please check again? Thank you! – Hermi Sep 17 '23 at 17:26
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    Good. Even clearer so: $(AB)^{-1}={c^{-1}\mid c\in AB}={(ab)^{-1}\mid a\in A,b\in B}={b^{-1}a^{-1}\mid a\in A,b\in B}=B^{-1}A^{-1},$ and (exchanging the roles of $A,B$) $(BA)^{-1}=A^{-1}B^{-1}.$ And again: showing $X\subset Y\iff X^{-1}\subset Y^{-1}$ is a waste of time: $\implies$ is sufficient. – Anne Bauval Sep 17 '23 at 17:32

1 Answers1

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One answer is reached through the comment section. I thought I would add my own solution here, by showing that if one of $AB\setminus BA$ and $BA\setminus AB$ is non-empty, then they both are non-empty.

Assume $AB\setminus BA$ is non-empty, and pick an element $x = ab\in AB\setminus BA$ with $a\in A$ and $b\in B$. Then consider $x^{-1} = b^{-1}a^{-1}$. Clearly it's an element of $BA$. Assume, for contradiction, that $x^{-1}\in AB$. Then it would be possible to write $x^{-1} = a_0b_0$ with some $a\in A, b\in B$. But then $$x = (x^{-1})^{-1} = (a_0b_0)^{-1} = b_0a_0\in BA$$which contradicts the choice of $x$.

So if there exists an $x\in AB\setminus BA$, then $x^{-1}\notin AB$, and therefore $x^{-1} = b^{-1}a^{-1}\in BA\setminus AB$. Therefore, if $AB\setminus BA$ is non-empty, then so is $BA\setminus AB$. By repeating the exact same proof, but with $A$ and $B$ swapped, this implication also goes the other way.

Arthur
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