I want to show this "iff" statement, but I already proved it, so I want to see if my proof is correct. Is the following proof true?

Caveat: $\textbf{For clarity and simplification, } a \cdot b = ab.$

Proof:

This proof uses $\underline{\textbf{Claim #1}}, \underline{\textbf{Claim #2}},$ and $ \underline{\textbf{Claim #3}}$ all togther.

Suppose that $G$ is a group. Then consider the following $\sim$ to be defined as $a \sim b\iff ab^{-1} \in G$ on the group $G$.

$\underline{\textbf{Claim #1:}} \sim$ is an equivalence relation.

$\textit{Proof: }$

Reflexive: If $a \in G$, then $a \sim a \iff aa^{-1} \in G \iff e \in G$, where $e$ is the identity element in $G$. Since $G$ is a group, $e \in G$. Thus, $a \sim a$.

Symmetric: If $a \sim b \Rightarrow ab^{-1} \in G$. But since $G$ is a group, $ba^{-1} = (ab^{-1})^{-1} \in G \Rightarrow b \sim a$.

Transitive: If $a \sim b$ and $b \sim c$, then $ab^{-1}$ and $bc^{-1}$ are in $G$. But G is a group. So, by closure, $(ab^{-1})(bc^{-1}) = a(bb^{-1})c^{-1} = aec^{-1} = (ae)c^{-1} = ac^{-1} \in G$. Thus, $a \sim c$.

Hence, $\sim$ is an equivalence relation. QED

==================================================================== Now, I want to show that $[a] = \{m \in G : m \sim a\} = G$. To show this, $[a] \subseteq G$ and $G \subseteq [a]$ is true, for all $a \in G$.

$\underline{\textbf{Claim #2:}}$ $[a] = G, \forall a \in G$

$\textit{Proof: }$

So, suppose that $x \in [a]$. Since $e \in [a]$ (because $ae^{-1} = ae = a \in G$), then $[a] = [e]$. So, $x \in [a] \iff x \in [e] \iff x \sim e \iff x = xe = xe^{-1} \in G \iff x \in G$.

$\therefore [a] \subseteq G$ and $G \subseteq [a]$ (by going backwards). QED

==================================================================== $\underline{\textbf{Claim #3:}}$ $a, b \in G \iff ab \in G$

$\textit{Proof: }$

To prove the "$\iff$" statement said in the title, first, let's start with the conditional.

1. That is, suppose that $a, b \in G$. Then, $ab \in G$ is trivial because $G$ is a group (by closure, one of the requirements to be a group).

Now with the converse:

2. Suppose that $ab \in G$. Then $ab \in G \Rightarrow a(b^{-1})^{-1} \in G \Rightarrow a \sim b^{-1} \Rightarrow a \in [a]$ and $b^{-1} \in [a] \Rightarrow a \in G$ and $(b^{-1})^{-1} = b \in G$ (because if $b^{-1} \in G$, then the inverse of $b^{-1}$ is in $G$). QED

Edit (Jan/05/22):

I just changed a few things for clarity. The reason is to make no more confusion.

Solution 1:

Here is an example which shows one of the problems addressed in the comments (see the title). Let $H=GL_n(K)$ the general linear group and $G=SL_n(K)$ be the subgroup of $H$, the special linear group. Then the diagonal matrices $$ a={\rm diag}(2,2\ldots ,2),\quad b={\rm diag}\left(\frac{1}{2},\frac{1}{2}\ldots ,\frac{1}{2}\right) $$ satisfy that $ab\in G$, but $a,b\not\in G$. So the title claim is incorrect.

Concerning your edit, where you said "I just changed a few things for clarity": you need to change the title and add this also to your post.