Suppose that $G$ is a group of order $30$ and has a Sylow $5$-subgroup that is not normal. Suppose that $G$ is a group of order $30$ and has a Sylow $5$-subgroup that is not normal. Find the number of elements of order $1$, order $2$, order $3$, and order $5$. But this scenario can't happen. Why not?
Let $G$ be a group of order $30=2 \cdot 3 \cdot 5$. 
The number of Sylow $2$-subgroup $n_2$ divides $15$ and has the form $n_2=2k+1$ by the Sylow theorems. Therefore $n_2=1,3,5,15$.
The number of Sylow $3$-subgroup $n_3$ divides $10$ and has the form $n_2=3k+1$ by the Sylow theorems. Therefore $n_3=1,10$.
The number of Sylow $5$-subgroup $n_5$ divides $6$ and has the form $n_5=5k+1$ by the Sylow theorems. Therefore $n_5=1,6$. However, since Sylow $5$-subgroup isn't normal $n_5 \neq 1$.
\begin{array}{|c|c|c|c|}
\hline
n_2 & n_3 & n_5 & number \,of \,elements & Possible \\ \hline
1  & 1  & 6 & 1+1\cdot2+6\cdot4=26     & Yes \\ \hline
1  & 10 & 6 & 1+10\cdot2+6\cdot4=45    & No  \\ \hline
3  & 1  & 6 & 3+1\cdot2+6\cdot4=29     & Yes \\ \hline
3  & 10 & 6 & 3+10\cdot2+6\cdot4=47    & No  \\ \hline
5  & 1  & 6 & 5+1\cdot2+6\cdot4=31     & No  \\ \hline
5  & 10 & 6 & 5+10\cdot2+6\cdot4=49    & No  \\ \hline
15 & 1  & 6 & 15+1\cdot2+6\cdot4=41    & No  \\ \hline
15 & 10 & 6 & 15+10\cdot2+6\cdot4=59   & No  \\ \hline
\end{array}
Where do I go from here?
 A: As discussed in the comments, the third row is the only possible option. Note that in this case the subgroup of order $3$ is normal. Thus the group has a quotient of order $10$. By Sylow considerations, a group of order $10$ has a normal subgroup of order $5$ and either a normal subgroup of order $2$, in which case the group is cyclic, or $5$ subgroups of order $2$, in which case there are $5$ elements of order $2$.
Suppose the quotient is cyclic. This would mean that the quotient has an element of order $10$, which would imply that the group of order $30$ has an element either of order $10$ or order $30$, but as all of the non-identity elements have been enumerated and have prime order this is not the case. 
If the quotient is not cyclic, it cannot have 5 subgroups of order $2$ because the group of order 30 only has three elements whose orders are even. Since these are the only two options for the quotient, there cannot be 6 Sylow 5-subgroups in a group of order 30, or in other words the Sylow 5-subgroup is normal.
A: My way: you deduced that for such a $G$ it must be $n_5=6$. These six subgroups are cyclic of order $5$ (every group of prime order is necessarely cyclic). Let $K_1,\dots,K_6$ be these $5$-Sylow. It must be $K_i\cap K_j=1$ for $i\neq j$ and every element different from the identity must has order 5. Hence $G$ contains exactly $6\cdot(5-1)=24$ elements of order $5$.
Again by your Sylow computation $n_3=1,10$. If was $n_3=10$ the by the same argument as above we'd have $10\cdot(3-1)=20$ elements of order $3$. So $G$ would contain more than $30$ elements. Hence $n_3=1$. So $G$ contains exactly $2$ elements of order $3$.
So till now we have $|G|=2\cdot3\cdot5$ hence it could contain only elements of order $2,3$ or $5$. And by Cauchy theorem it must contain at least one element of order $2,3$ and $5$.
Moreover we counted (togheter with the identity element which is the only element of order $1$) $24+2+1=27$ elements in $G$. The three remaining elements must be of order $2$, there is no other way (and from this we deduce $n_2=3$).
Summing up, we have that if $G$ is a group of order $30$ with a $5$-Sylow not normal, it must be:
$\bullet$ $n_2=3,\;\;n_3=1,\;\;n_5=6$;
$\bullet$ $G$ has exactly $1$ element of order $1$, $3$ elements of order $2$, $2$ elements of order $3$ and $24$ elements of order $5$.
