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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?

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I suppose that after "proof" you actually meant to write "Suppose G is abelian"...then erase that "not". – DonAntonio Oct 27 '12 at 11:25
yes that is What I wanted to write. Thanks. – Reader Oct 27 '12 at 11:34
But still if I wanted to complete the argument above, how would I do it? – Reader Oct 27 '12 at 11:40
I can't see how since you could have chosen $\,theta\,$ a generator of the field extension and thus you have no contradiction at all... – DonAntonio Oct 27 '12 at 11:53
Do I have that $[\mathbb Q(\theta):\mathbb Q]=5$ so the order of the group G is 5, which is a contradiction? Is it correct? – Reader Oct 27 '12 at 12:37
up vote 4 down vote accepted

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 $10$.

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+1 Nice and short. – DonAntonio Oct 27 '12 at 11:30
Nice and easy. Thanks. – Reader Oct 27 '12 at 11:39
One thing I dont understand is: S_5 has no element of order 10 and G has element of order 10, then what? How do we have that G cannot be the Galois group? – Reader Oct 27 '12 at 13:05
How does it follows that G is not abelian. – Reader Oct 27 '12 at 14:41
@Reader The only abelian group of order 10 is cyclic (see this with the structure theorem or prove it directly) and so it has an element of order 10. On the other hand, $G$ is a subgroup of $S_5$: it's essentially made up of certain permutations of the zeros of $f$. If $G$ were cyclic, then $S_5$ would have to contain an element of order $10$ as well, and it doesn't. – Cocopuffs Oct 27 '12 at 15:04

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