Possible Error in Proving the One-Step Subgroup Test

Question

The One-Step Subgroup Test is as follows:

"Let $G$ be a group and $H⊆G$ be a subset of $G$. Then $H≤G$ if, and only if, $H$ is non-empty and $ab^{−1}∈H$ for all $a,b∈H$."

So I had asked a question before about this one, but this time, I'm now asking for some verification on proving the converse of the statement instead of proving the forward direction. In other words, showing that $H$ would be a subgroup as a result.

In one textbook I'm reading, the proof is as follows:

"Since $H \neq \emptyset$, there exists an element $b \in H$ for which $bb^{-1} \in H$. But $bb^{-1} = 1_G \in H$, so therefore $1_G$ is the identity."

One of the main problems I see in this proof is that it doesn't show that $b^{-1}b = 1_G$ in $H$, and you can't automatically assume that it is since H is not necessarily commutative. So how would $1_G$ be the identity of $H$ then? Is there something I'm missing here?

Thanks in advance.

Answer 1

$b^{-1}b=b^{-1}(b^{-1})^{-1}=(bb^{-1})^{-1}=e^{-1}=e$, by the so-called "socks and shoes property".

Answer 2

A basic result in group theory:

If $ab=1$ then $ba=1$ (so $b=a^{-1}$).

Proof: $ba=ba(bb^{-1})=b(ab)b^{-1}=bb^{-1}=1$.