Prove that a group of order $2^{13}\cdot 13$ has a proper normal subgroup.
Let $n_2$ be the number of sylow $2$ subgroups and $n_{13}$ the number of $13$ sylow subgroups. By the sylow theorems have $n_2 \mid 13$ and $n_2 \equiv 1 \mod 2$ and $n_{13}\mid 2^{13}$ with $n_{13} \equiv 1 \mod 13$.
Claim either $n_2=1$ or $n_{13}=1$, assume by contradiction that not the case, then we have $n_2 = 13$ and $n_{13} = 2^{12}$. Counting elements since $13$ sylow subgroups cyclic they have trivial intersection. Then we have $12(2^{12}) = 12 \cdot 2^{12}$ distinct elements of order $13$, on the other hand each $2$ sylow subgroup contains $2^{13}$ elements, we cannot say that they have trivial intersection right? Then how to proceed?
Thanks
You need to show your group $G$ is not simple. Assume it is simple. Then $n_2\ne 1$, and so $n_2=13$. Define an action of $G$ on the collection of $2$-Sylow subgroups $Syl_2(G)$ by conjugation: $g.P=gPg^{-1}$. The action induces a group homomorphism:
$\varphi:G\to S_{Syl_2(G)}\cong S_{13}$
by $\varphi(g)(P)=g.P=gPg^{-1}$. Since $G$ is simple by assumption, $\varphi$ must be either trivial or injective. It can't be injective because $|G|=2^{13}$ does not divide $13!=|S_{13}|$. And it also can't be trivial. Indeed, let $P,Q$ be two distinct $2$-Sylow subgroups. By the Sylow theorems they are conjugate, i.e there is some $g\in G$ such that $gPg^{-1}=Q$. Then this $g$ satisfies:
$\varphi(g)(P)=gPg^{-1}=Q\ne P$.
And so $\varphi(g)\ne id$. So $\varphi$ is neither trivial nor injective, a contradiction.