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Let $G$ be a multiplicative group and $A\subseteq G$ such that

1) $\forall a\in A, a^{-1}\in A$

2) $1\in A$

3) $AA \subseteq gA$ for some $g\in G$

Can we say that $A$ is a subgroup?

One can immediately show that $g\in A$ and that we have $g^{-1}A \subseteq A \subseteq gA$ but I believe you must use the symmetry property of $A$ (1) again to conclude that it is a subgroup (if it is).

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up vote 8 down vote accepted

$1=1\cdot 1\in AA\subset gA$ so $1=ga_0$ for some $a_0\in A$. Since $a_0=g^{-1}$, we have $g^{-1}\in A$ hence by 1) $g\in A$. We have $gA=A$. Indeed, $A=1\cdot A\subset A\cdot A\subset gA$ and $g^{-1}A\subset gA$ so $A\subset g^2A$. In particular, $1=g^2a$ so $(g^2)^{-1}\in A$ and $g^2\in A$. So $g^2A\subset A\cdot A\subset gA$ and if $x\in gA$ then $x=ga$ for some $a\in A$ so $gx=g^2a\in g^2A\subset gA$ hence $gx=ga'$ for some $a'\in A$ hence $x\in A$ and $gA=A$.

So $A\cdot A\subset gA=A$, and with 1) and 2) it shows that $A$ is a subgroup of $G$.

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I'm sorry, but I do not quite get how $gA= A$ since $A$ is just a subset. – Santiago C. Jan 25 '12 at 18:49
Indeed, it was not clear and I think it's the main part of the proof. I have added the details. – Davide Giraudo Jan 25 '12 at 19:11
Surely the final comment about "normal subgroup" can't be right-- what if in fact $g=1$? (Then trivially $A$ is a subgroup, but it could be any subgroup). – Matthew Daws Jan 26 '12 at 20:53
@matthew Oh yes you are perfectly right. I will remove it. I am asking myself why I wrote that. – Davide Giraudo Jan 26 '12 at 20:55
@DavideGiraudo: I have no idea either ;-) – Myself Jan 26 '12 at 23:11

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