If $|G|=3^3 13$ then $G$ has a non-trivial normal subgroup If $|G|=3^3 13$ prove that $G$ has a non-trivial normal subgroup

since $13$ is prime that divides $|G|$ from Cauchy's theorem we get that it exist a subgroup with order $13$, because the order is prime the group is cyclic and Abelian, thus it's a normal subgroup.
The solution for this exercise was using the Sylow theorems to prove this, which is way more time-consuming, is my answer wrong ?
 A: Your solution is incorrect. For example in $S_{3}$ . The subgroup $\{(1),(12)\}$ is abelian as it is of order $2$ but it is NOT normal in $S_{3}$.
The correct solution is :-
Let $n_{13}$ denote the number of Sylow-13 Subgroups .
Then from Sylow's third theorem .
$n_{13} = 13k+1$ for some non-negative integer $k$.
And $n_{13}| 27$.
So $k=0$ or $1$.
If $n_{13}=1$ . Then by again by uniqueness of the subgroup you have that it is normal(By sylow's theorem the sylow subgroups of a particular order are conjugates) and hence it will prove our claim.
If $n_{13}=27$ and $n_{3}=1$ . Then again the unique subgroup of order $27$ is normal and it would prove the claim.
The only case that requires special treatment is $n_{13}=27$ and $n_{3}=13$. But this is simply not possible. As You get $12\cdot 27=324$ elements of order $13$.
And you are left with atmost $27$ more elements. But you cannot fit them inside $13$ subgroups of order $27$. Hence you must have only $1$ subgroup of order $27$. Which would be normal by uniqueness .
Hence proved.
A: I want to show that the number of $Syl_{13}(G)=1$
From the $1st$ theorem, it exists such subgroup.
From the $3nd$ theorem $n_{13}=\{1,27\}$
If $n_{13}=1$ we're done.
If  $n_{13}=27$ then there are $27$ subgroups of order $13$ that means $27\times 12+1=325$ elements.
$G$ also have groups with orders $3,9,27$ that cannot be contained in the above subgroups since are relative primes. Then we will have $2+8+26+1$ additional elements, which make a total of $362> |G|$
