Every group of order $14077=7 \cdot 2011$ is cyclic (use Sylow Theorem)

Question

I know that I can use Sylow's theorem because $7$ is prime, and so is $2011$. So, $14077$ has the form $p^{a}m$ where $p \nmid m$ for some $a$ and $p$ is prime. Moreover, we know from previous results that there are exactly two subgroups of the group of order $14077$, one subgroup for each divisor of $14077$. Both of those subgroups then, call them $Q_{7}$ and $Q_{2011}$ would be cyclic because each has prime order. So...is this good enough to prove that the group of order? I feel like I am missing a step, and I have seen other problems use external direct products. Do I need to do that here to show $Q_{14077}=Q_{7}$ X $Q_{2011}$?

Answer 1

Let $G$ be a group such that $|G| = 14077 = 7 \cdot 11$. Since for any prime $p$ the number of Sylow$_p$ subgroups is congrue to $1$ mod$p$ we know there is one Sylow$_{2011}$ and hence a normal Sylow$_{2011}$ subgroup. Let $P$ we the unique Sylow$_{2011}$ group.

Now since the normalizer of a subgroup mod the centralizer of that subgroup is isomorphic to a subgroup of the auotmorphism group of that subgroup we have $N_G(P)/C_G(P) = G/C_G(P) \cong H$, where $H$ is a subgroup of Aut$(P)$. Since $2011$ is prime we have that $P$ is cyclic and hence $|$Aut$P| = \phi(2011) = 2010$. Since gcd$(|G|, 2010) = 1)$ we get $|H| = 1$. Hence $P \in Z(G)$. But $|G/P| = 7$ which is prime and hence cyclic. Therefore $G$ mod a subgroup of the center of $G$ is cyclic so $G$ is abelian. So every subgroup of $G$ is normal.

Let $K$ be a normal and hence unique Sylow$_7$ subgroup. Since $G/P$ is cyclic and in particular prime $G' < P$. Since $|G/K| = 2011$ we get that $G/K$ is also abelian so $G' < K$. Since $P \cap K = 1$ we have $G' = 1$. Therefore $G$ is abelian. From the fundamental theorem of finite ableian groups we have that $G$ is cyclic.