Left Equivalence Relation I need to prove that if $G$ is a group, $H$ is a subgroup of $G$ and $\equiv_{L}$ is the equivalence relation from the partition $\{xH\:|\:x\in G\}$, then for $x_{1},x_{2}\in G$, $x_{1}\equiv_{L}x_{2}\Leftrightarrow x_{1}^{-1}x_{2}\in H$.

Attempt at Proof: Assume $x_{1}\equiv_{L} x_{2}$. By definition, 

$x_{1}\equiv_{L} x_{2} \iff x_{1}H=x_{2}H$

$\iff H=x_{1}^{-1}x_{2}H$.

The last "if and only if" statement happens only when $\overline{x_{1}^{-1}x_{2}}=\overline{1_{G}}$, where the overline denotes an equivalence class. But $\overline{x_{1}^{-1}x_{2}}=\overline{1_{G}} \iff x_{1}^{-1}x_{2}\in H$. Hence, $x_{1}\equiv_{L}x_{2}\Leftrightarrow x_{1}^{-1}x_{2}\in H$.

Does this proof look okay? I think it is logical, but I don't want to make any mistakes on such a trivial problem. Thanks in advance for any help!
If you are using the partition $\{xH\mid x\in G\}$ to define your equivalence relation, then by definition $x_1\equiv_L x_2$ if and only if $x_2\in x_1H$. Therefore, the proof would be:
\begin{align}
x_1\equiv_L x_2 &amp;\iff x_2\in x_1H\\
&amp;\iff x_2=x_1h\mbox{ for some }h\in H\\
&amp;\iff x_1^{-1}x_2=h\in H.
\end{align}
If you're doing a step-by-step 


A $\Leftrightarrow$ B 

B $\Leftrightarrow$ C

etc. 

proof, you don't need to assume A at the beginning.


Your proof is essentially structured as claiming A $\Leftrightarrow$ B and B $\Leftrightarrow$ C and then going back and saying D $\rightarrow$ (B $\Leftrightarrow$ C) and D $\Leftrightarrow$ E, which a confusing structure.
It might be good to just work out why $x_{1}H=x_{2}H\iff H=x_{1}^{-1}x_{2}H$ from the definition of $x_{1}H$. On high level proofs, one can assume simpler properties that have already been established in the course, but if you're doing a basic proof involving X, you should at some point be relying on the definition of X.