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

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?

Shaun
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mathematics-and-caffeine
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1 Answers1

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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.

markvs
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    What do you mean by class here? – Derek Holt Feb 13 '22 at 14:48
  • @DerekHolt Solvable groups of class $k$ or soluble groups of derived length $k$ or разрешимые группы класса $k$, if you prefer. – markvs Feb 13 '22 at 15:24
  • For example https://mathoverflow.net/questions/229505/finitely-presented-subgroup-and-free-solvable-group-of-class-3 – markvs Feb 13 '22 at 15:51
  • or: http://math.caltech.edu/~jaywill/WeaklyUniversalFGSolvable.pdf or https://web.stevens.edu/algebraic/GTI/Files/2011-11-03-talk-Romankov.pdf or https://math.stackexchange.com/questions/185591/an-example-of-a-2-generated-nonlinear-solvabe-group or https://arxiv.org/abs/1204.6506

    or http://pi.math.cornell.edu/~tim.riley/papers/Taming_the_hydra/taming_the_hydra_extended_abstract.pdf

     – markvs Feb 13 '22 at 16:05