Let $G$ be a group of order $2014$. Prove that $G$ has a normal subgroup of order $19$ and $G$ is solvable

Question

Let $G$ be a group of order $2014$. Prove that $G$ has a normal subgroup of order $19$ and $G$ is solvable.

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The first part directly follows from the Sylow Theorems, if you write $2014 = 2 \cdot 19 \cdot 53$.

But I really don't know how to prove that it is solvable. How to prove that?

Answer 1

If $S$ is the Sylow $19$-subgroup then $S$ is normal and cyclic, $|G/S|=2*53$. $G/S$ has unique cyclic Sylow $53$-subgroup $T/S$ of index $2$, so $G$ has a sequence $1<S<T<G$ of normal subgroups with Abelian factors. Hence $G$ is solvable of class at most $3$. In fact the class is $2$ or $1$ because $T$ is a direct product of two groups of orders $19, 53$, hence cyclic.