Closed Property: Basic Subgroup Definition

Question

Let $*$ denote a operation. Also, let $S$ be a non-empty subset of a group, namely $G$.

We say $S$ is a subgroup of G: if $a,b \in S$, then $a* b^{-1}\in S$.

Why does this definition imply if $a,b\in S$, then $a*b\in S$?

I feel like I am missing something simple, but I have no idea where to go from here.

Answer 1

Take $a=b$, then it follows that $e\in S$.

Then take $a=e$, then it follows that $b^{-1}\in S$.

Now take $b^{-1}$ instead of $b$ and get the desired result.

Answer 2

So $e=ab^{-1}(ab^{-1})^{-1}\in S$. So $b^{-1}=eb^{-1}\in S$ and hence $ab=a(b^{-1})^{-1}\in S$.

Answer 3

We have:

$\forall a, b \in S \; ab^{-1} \in S; \tag 1$

let

$c \in S; \tag 2$

we know that such $c$ exists since $S \ne \emptyset$; taking

$a = b = c \tag 3$

in (1) we find

$e =cc^{-1} \in S, \tag 4$

where $e \in G$ is the identity element. Thus, again using (1)

$b \in S \Longrightarrow b^{-1} = eb^{-1} \in S; \tag 5$

thus

$a, b \in S \Longrightarrow a, b^{-1} \in S \Longrightarrow ab = a(b^{-1})^{-1} \in S, \tag 6$

and we are done!