Proving $H\triangleleft G$ for a finite group $G$ and a subgroup $H$ Let $G$ be a finite group of odd order and $H$ a subgroup of index $3$ or $5$. 
How to prove, that $H\triangleleft G$?
 A: Consider the action of $G$ on the set of left cosets of $H$. That is we have a homomorphism onto $S_3$ or $S_5$ respectively (symmetric group on 3 or 5 elements). Suppose first we are in the $S_3$ case (i.e. $H$ is of index $3$). Let $K$ be the kernel of the above homomorphism which is contained in $H$. After some thinking (and using that the order of $G$ is odd) you will obtain that $G/K$ has order $3$. Use the fact that $[G : K] = [G : H] [H : K]$ and obtain that $[H : K] = 1$. The $S_5$ case is a bit different but you should also be able to show that $K = H$ in this case and thus that $H$ is normal.
A: In case $H$ is a subgroup of index $3$, we use the following theorem: let $G$ is a finite group. Let $p$ be the smallest prime divided by $\#G$. If $H$ is a subgroup of $G$ and index of $H$ is $p$, then $H$ is normal. (the proof is exactly the same as below, but much more simple as you will only have $\#\phi(G)=p$)
Apply this in the case where $G$ is odd, then $3$ is the smallest prime which divided by $\#G$. So we have $H$ is normal.
In case $H$ is a subgroup of index $5$, consider the homomorphism: 
$$\phi: G \mapsto Bij(G/H) \cong S_5$$
We have: $\#\phi(G)$ is divided by both $\#G$ and $\#Bij(G/H)=\#S_5=120$. Since $\#G$ is odd, we must have $\#\phi(G)\in \{3,5,15\}$. We will prove that only $5$ is possible.


*

*If $\#\phi(G)=3 \Rightarrow \#\ker(\phi)=\frac{\#G}{\#\phi(G)}=\frac{\#G}{3} > \frac{\#G}{5}= \#H$.
On the other hand, it is easy to show that $\ker(\phi) \subset H \Rightarrow$ contracdiction 

*If $\#\phi(G)=15$. We will show that, there doesn't exist a subgroup of $S_5$ whose index is $15$. Suppose $K$ is a subgroup of $S_5$ whose index is $15=3\times 5$. By Cauchy, there must exist at least $1$ element of order $3$ and $1$ element of order $5$ in $K$, thus a $3-$cycle and a $5-$cycle, say $a$ and $b$, respectively. Clearly, they do NOT have disjoint support, thus they are NOT commutative.
We will generate $K$ based on these $2$ elements. Their inverses must also be in $K$. So $K$ now has $5$ elements: $e,a,a^{-1},b,b^{-1}$. 
So the above set can generate $8$ more different elements. Amongst those, $ba$ and $a^{-1}b$ generate $2$ new different elements: $b^2$ and its inverse $b^{-2}$. But if so, $ab^2,a^{-1}b^2,a^{-1}b^{-2},a^{-1}b^2$ are also $4$ new different elements. So now, $K$ has at least $19$ elements $\Rightarrow$contradiction

*So we must have $\#\phi(G)=5 \Rightarrow \#\ker(\phi)=\frac{\#G}{5}= \#H$. Combining this with the fact that $\ker(\phi) \subset H$, we conclude: $H=\ker(\phi)$, thus normal
