Prove that $A$ is a subgroup of the group $G$ if and only if $AA^{−1} ⊆ A$. If $A$ is a subset of a group $G$, we let $A^{−1} = \{a^{−1} : a \in A\}.$ Also, for $A,B \subseteq G$, we let $AB = \{ab : a \in A,b \in B\}$. Prove that $A$ is a subgroup of the group $G$ if and only if $AA^{−1} \subseteq A$.
The only conclusion I can see is that the identity element is contained within $A$. The subgroup tests are probably necessary, but I am having difficulties making the connection. Any help would be appreciated. 
 A: It appears that the more difficult direction is to show that $AA^{-1}\subseteq A$ implies $A$ is a subgroup. I'm gonna skip the identity element part, because you said you understand that. So let's say we've already established that $e\in A$.
Next let's prove that $A$ is closed under taking inverses. For any $x\in A$, $x^{-1}\in A^{-1}$. Also using the already established fact that $e\in A$, we can see that $x^{-1}=ex^{-1}\in AA^{-1}\subseteq A$, i.e. $x^{-1}\in A$.
Finally, let's show that $A$ is closed under multiplication. Pick any $x,y\in A$. As we established above, $y\in A$ implies $y^{-1}\in A$, and therefore $y=(y^{-1})^{-1}\in A^{-1}$. Thus $xy\in AA^{-1}\subseteq A$, i.e. $xy\in A$.
It seems to me that you didn't need help with the other direction — that if $A$ is a subgroup then $AA^{-1}\subseteq A$. But I can add this part too, if necessary.
A: $A$ is a group iff it is closed under inversion, multiplication (which implies it contains the multiplicative inverse) iff $A^{-1}$ is a subset of $A$ and $AA$ is a subset of $A$. But $A^{-1}$ being a subset of $A$ is equivalent to $A^{-1}=A$. Thus we have $A$ is a group iff $A^{-1}=A$ and $AA$ is a subset of $A$ iff $A^{-1}=A$ and $(A^-1)A$ is a subset of $A$. Thus we only remain to show $A^{-1}=A$ is necessary for $A^{-1}A$ to be a subset of $A$. But when $A^{-1}A$ is a subset of $A$, in particular $\{aa^{-1}\}=\{e\}$ is a subset of $A$ and thus in particular $A^{-1}{e}$ is a subset of $A$ and $A^{-1}=A$. 
