Let $G$ be a finite group and $H\triangleleft G$ a normal subgroup. Prove that $|G/H| =|G|$ if, and only if, $H = \{e\}$.

As @Jacob noted above, the proof is very short and it looks so clear. Since $H$ is a normal subgroup of $G$, so we can speak about a new group called Quotient Group $G/H$ which contains all right cosets of $H$ in the group. According to the definition, the index of $H$ in $G$ is denoted by $[G:H]$ and if the mother group is finite, you are allowed to write it as $$[G:H]=\frac{|G|}{|H|}$$ Here, $G$ is finite, so $|G|<\infty$ and then we have $$[G:H]=\frac{|G|}{|H|}=|G|\Longleftrightarrow H\cong\{e\}$$

Hints:

You need to know your notation and definitions, that's about all there is to it.

Proofs are usually first and foremost about unpacking definitions, and knowing the notation.

What does it mean for $H \triangleleft G$ to be normal in $G$?

• Answer: Yes, confirming your answer below,
  
  "A subgroup H of a group G is normal in G if and only if the left cosets equals the right cosets of $H$ in $G$ i.e. $gH=Hg$.

What does $[G:H]$, the index of $H \in G$, represent? And what does this mean with respect to $|G/H|$?

• "... So the index represents the total number of cosets of $H$ in $G$."
  
  Yes, since $H$ is a normal subgroup of $G$, we can form the quotient group $G/H$ which contains all cosets of $H$ in the group $G$, so $[G: H] = |G/H|.$

And what must follow if $|G/H| = |G|$? I.e. What must the order of $H$ be for $|G/H| = |G|$ to be true?

• "... the order of $H$ should be $1$ for the equation to hold.":
  Yes indeed.

And what is the only normal subgroup of $G$ of that order (it's the only subgroup, period, of that order)?

• "...so that means the only normal subgroup in $H$ is $H$ itself?"
  
  Well, yes, $H$ is normal, as given: it means the only normal subgroups $H \triangleleft G$ such that $|G/H|= |G| \implies |H| = 1\;$ is $H = \{e\}$, the trivial (normal) subgroup.