Question: Let $G$ be an finite group and $H$ normal subgroup of $G$. Show $\left|G\big/H\right|=\left|G\right|$ if and only if $H=\{e\}$.

Firstly I do not know how to show that $G$ is finite. Next I know that if $H$ is a normal subgroup of $G$ then for all $a \in G$ I have the cosets $aH = Ha$.

Please help.

Answer: 
You are given a finite group $G$ and a normal subgroup $H$ of $G$, so you don't need to prove that $G$ is finite and that $H$ is normal.

Regarding the statement 

> 
> $\left|G\big/H\right|=\color{blue}{|G|}\quad$ if and only if $\quad H=\{e\}$
> 
> 
> 

do you know Lagrange's theorem?

(Note: In your original question you wrote $\left|G\big/H\right|=\color{red}{\left|H\right|}$, I'm pretty sure that was a mistake.)
*Hint*: the cosets of $H$ in $G$ are disjoint, all have the same size, and the union of them is $G$. Keep in mind that $H$ itself is a coset of $H$ in $G$. $|G/H|$ is the number of cosets of $H$ in $G$. If $|G/H|=|G|$, how big can the cosets be?