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?
Best Answer
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.