Show a group of order 351 is NOT simple. I have read a bunch of answers around the web, but they perform a jump which i can't follow.
I have determined that there must be $12*27=324$ elements of order 13 in G, but when i try to count the amount of elements in G of order 3 i run into some problems, i don't get the contradiction i was hoping for.
Some answers i found on the web, just says that since there is $3^3=27$ elements left in G, these must amount to a unique and therefore normal subgroup in G.
But why is it so?
My proof so far:
Since $|G|=351=3^3*13$ Sylow theorem force $n_3$ to be $1$ or $13$, and $n_{13}$ to be $1$ or $27$. Show that neither $n_3$ nor $n_{13}$ is $1$, for contradiction. By lagrange theorem the order of the elements in the Sylow 13-subgroup must be $1$ or $13$, the only element with order $1$ is the $e$, which means there is $12$ elements, in each of the $27$ Sylow 13-subgroups, of order 13. Which means here are $12*27=324$ elements of order 13 in $G$. There remains $351-324=27=3^3$ elements in $G$, which must amount to a Sylow 3-subgroup, because?
I think i really need some Explain it like im 5 type of stuff here.

If $n_3=1$, we are done. Suppose not, so there are 13 Sylows for 3. If there is one 13 Sylow there is nothing to prove, so suppose also there are 27 Sylows for 13. If $P$, $Q$ are two 13-subgroups then they are either disjoint or equal for every nontrivial element in them is a generator, since 13 is prime. This justifies the statment there are $27\cdot 12$ elements of order 13. This leaves 27 elements in $G$ not of order 13. In particular there are at most 27 elements with order a power of 3, contradicting the fact there are 13 distinct Sylow 3-subgroups.
