Show that any group of order 294 is solvable.

group-theorysolvable-groupssylow-theory

Show that any group of order 294 is solvable.

Proof: Suppose $G$ is a group of order $294=2\cdot 3\cdot 7^2$. By Sylow's Theorem,

  • $n_2\equiv 1 \mod 2$ and $n_2|147$ $\Rightarrow$ $n_2=1,3,7,21,49,147$.
  • $n_3\equiv 1 \mod 3$ and $n_3|98$ $\Rightarrow$ $n_3=1,7$.
  • $n_7\equiv 1 \mod 7$ and $n_7|6$ $\Rightarrow$ $n_7=1$.

Hence there is a normal Sylow $7$-subgroup, $H$, with order $49$.

So where am I suppose to go from here?

Best Answer

The normal Sylow $7$-subgroup $H$, being a $p$-group, is solvable. The quotient $G/H$ has order $6$ and is thus solvable (the first non-solvable group has order $60$). Hence $G$ is solvable.

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