Galois theory (Showing $G$ is not abelain) Suppose $G$ is the Galois group of an irreducible degree $5$ polynomial $f \in \mathbb{Q}[x]$ such that $|G| = 10$. Then $G$ is non-abelian. 
Proof: Suppose $G$ is abelian. Let $M$ be the splitting field of $f$. Let $\theta$ be a root of $f$. Consider $\mathbb{Q}(\theta) \subseteq M$. Since $G$ is abelian every subgroup is normal. This means $\mathbb{Q}(\theta) \subseteq M$ is a normal extension.  So $f$ splits completely in $\mathbb{Q}(\theta)$. Then what how to complete the proof. How would I get a contradiction?

The only abelian group of order $10$ is cyclic. Since $G$ is a subgroup of $S_5$, it's enough to show that there's no element of order $10$ in $S_5$.
If you decompose a permutation in $S_5$ as a product of disjoint cycles, then the order is the LCM of the cycle lengths - and these can be any partition of $5$.
Since $5 = 1 + 4 = 1 + 1 + 3 = 2 + 3 = 1 + 1 + 1 + 2 = 1 + 2 + 2 = 1
+ 1 + 1 + 1 + 1$ are the only partitions, the only orders that appear are $1,2,3,4,5,6$ and in particular not $10$.
