# Galois group is isomorphic to $S_5$?

Let f be an irreducible polynomial of degree $5$ in $\mathbb{Q}[x]$. Suppose that in $\mathbb{C}$, $f$ has exactly two nonreal roots. Then the Galois group of the splitting field of $f$ is isomorphic to $S_5$.

My effort: Let $G$ be the Galois group. The complex conjugation map $\mathbb{\sigma} (x+iy)=x-iy$ is non-trivial element of $G$ of order $2$. Let $\alpha_1,\alpha_2, \alpha_3$ be the real roots of $f(x)$ and $\beta_1$ and $\beta_2$ be the nonreal roots. Then, $\sigma$ sends $\beta_1$ to $\beta_2$ while keeping other roots fixed. Now if we can construct an element of order $5$ in $G$, then we can proceed in the direction of showing that $G \cong S_5$.

I know that any such element $\tau$ of $G$ (if exists) will permute all the roots of $f$. But I am not sure which permutation of roots of $f$ will give me the correct candidate for the element $\tau$?

Thanks!

• What's the question? – Lord Shark the Unknown Jul 14 '18 at 16:41
• Demanding others solve a problem rather than answer an actual question makes it look like this is a homework problem. In fact it is a standard type of problem for students learning Galois theory. – KCd Jul 14 '18 at 16:48
• @KCd please have a look. I have put my efforts along with question. – Shubham Namdeo Jul 14 '18 at 16:57
• The Galois group of an irreducible polynomial over a field of characteristic zero will act transitively on its zeros. – Lord Shark the Unknown Jul 14 '18 at 17:01
• @LordSharktheUnknown after that how to proceed? – Shubham Namdeo Jul 14 '18 at 17:04

Proof 1(Direct approach)

Theorem Let $$p$$ be prime. If a transitive subgroup $$H$$ of $$S_p$$ contains a transposition $$(1 \ 2)$$, then $$H=S_p$$.

Define a relation $$\sim$$ on $$\{1,2,\ldots, p\}$$ by $$a \sim b \ \Longleftrightarrow (a \ b)\in H.$$

This relation is an equivalence relation (reflexive, symmetric, transitive).

Let $$[a]_{\sim}$$ be the equivalence class containing $$a$$. Since $$(1 \ 2)\in H$$, we have $$1 \sim 2$$.

Since $$H$$ is transitive, we can prove that there is a bijection between any two equivalence classes.

Then the cardinality of $$_{\sim}$$ must divide $$p$$. Since $$\{1,2\}\subseteq _{\sim}$$, we must have $$|_{\sim}|=p$$. This tells us that all transpositions containing $$1$$ are in $$H$$. Hence, $$H=S_p$$.

Proof 2(Galois Theory)

If $$G$$ is Galois group of an irreducible polynomial $$f(x)\in\mathbb{Q}[x]$$ of degree $$p$$, then $$p$$ divides the order of $$G$$. Also, $$G$$ is transitive and it contains a transposition. Since $$G$$ must have an element of order $$p$$, it contains a $$p$$-cycle. Hence $$G\simeq S_p$$.