Let $f = x^5 + x^4 + x^3 -2 x^2 + x + 1$. Does $\text{Gal}(f)$ equal to $A_5$?

By calculating the discriminant of $f$, I get $\text{Disc}(f)=42849=207^2$. So $\text{Gal}(f)$ is a subgroup of $A_5$. Also, since $5$ divides $|\text{Gal}(f)|$, $|\text{Gal}(f)|$ could be $5,15,20,60$.

I also noticed that $\text{Gal}(f)$ should have a 3-cycle. The reason is that $f$ modulo $3$ has exactly two roots: $\bar{1},\bar{2}$. As a result, $\bar{f} = (x^3 + x^2 + 2x + 2)(x+2)(x+1)$.

So that rules out $5$ and $20$. How should I proceed from here?


Actually, there is no subgroup of order 15 in $S_{5}$. (See here)

To prove this, suppose there exists $H\leq S_{5}$ with $|H|=15$. Since every group with order $15$ is cyclic, $H=\langle \sigma\rangle$ for some $\sigma\in S_{5}$. But if we consider cycle decomposition of $\sigma$, it's order can't be 15.


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