# Showing a subset is a subgroup

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.

• 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 $a_1,a_2.$ – Thomas Andrews Apr 23 at 15:40
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, $$ab\in aA=A$$.
• 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? – JustWandering Apr 24 at 14:27