Let $G$ be a simple group. Show that if $H$ is a subgroup $G$ such that $[G:H]= 3$ then $H=\{1\}$ and $|G|=3$. MY ANSWER: Since $G$ is simple the only normal subgroups of $G$ are $\{1 \} $ is $ G $ itself. As $ G / H $ is a quotient group it follows that $ |H|\neq 3 $ because otherwise the $ H $ index in $ G $ would be 1. Therefore, $ H = \{1 \} $ and therefore $ | G | = 3 $. 
Can you check if my answer is right?
 A: Your argument is thoroughly incorrect (and almost incoherent, to be honest). You do not know that $H$ is normal, you are only told it has index $3$ and is a subgroup. So you cannot assert that $G/H$ is even a group. Moreover, you cannot assert that $|H|\neq 3$, and your reasoning does not make sense; if $|H|$ were equal to $3$, that would not imply that the index is $1$... at least not immediately; what makes you think it does? It is certainly possible for a subgroup to have the same order as its index, and the order and the index of a subgroup need not be related at all.  All you know is that the subgroup has index $3$.
Instead, note that the action of $G$ on the left cosets of $H$ induces a homomorphism $G\to S_3$ whose kernel is a normal subgroup contained in $H$, and since $G$ is assumed to be simple, the kernel must be trivial. Can you take it from there?
A: You probably meant that there's a normal subgroup of index $3$.  Then you would be correct, if you would note that $H$ can't be a nontrivial proper subgroup.  But it would be if it had index $3$, unless...
In the problem it's not actually necessary to assume $H$ is normal, as @Arturo's solution shows.
A: On page 253 in this, we see that "Corollary 12.6. Let $G$ be a finite group, and let $H \leq G.$ Assume that $|G : H| = p,$
and let $p$ be the smallest prime divisor of $|G|$. Then $H\unlhd G.$ And your question deduces $p=3.$
