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Let $A$ be a subset of a finite group $G$. For any $g\in G$ we denote by $gA$ the set $\{ga : a ∈ A\}$. Assume that $e \in A$, where $e$ is the identity element of $G$, and that for any $g_1, g_2 \in G$ the sets $g_1A$ and $g_2A$ are either equal or disjoint. Show that $A$ is a subgroup of $G$.

My attempt

Use subspace test. Firstly, we already know that $e$ is in $A.$

Suppose $g_1A=g_2A$ then there exists some $a_1,a\in A$ such that $g_1*a_1=g_2*a_2$ then $g_1*e=g_1=g_2*a_2*a_1^{-1}\in gA$ which implies that $a_2*a_1^{-1}$ is an element in set A and so A is a subgroup by subgroup test.

My problem

How should I approach the disjoint part? I thought about using equivalence relations since this is essentially a partition but I am not entirely sure how to continue.

Thanks very much in advance.

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  • $\begingroup$ You haven't proved that $a_2a_1^{-1}\in A$ for all pairs $a_1,a_2\in A.$ You've only showed it for specific values of $a_1,a_2.$ $\endgroup$ – Thomas Andrews Apr 23 at 15:40
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As $G$ is finite, you might proceed by showing taht $A$ is not empty and $A$ is closed under multiplication. $A\ne\emptyset$ follows from $e\in A$. For closure, let $a,b\in A$. As $a=ae\in aA$, we see that $aA$ and $A=eA$ are not disjoint, hence are equal. In particular, $ab\in aA=A$.

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  • $\begingroup$ Would we potentially have to check the closure under inverses as well? Because the subgroup test states $ab^{-1}∈A$ right? Or is this unnecessary? $\endgroup$ – JustWandering Apr 24 at 14:27

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