Suppose $G$ is the Galois group of an irreducible degree 5 polynomial $f\in\mathbb{Q}[x]$ such that $|G|=10$. Then $f$ must have precisely 1 or 5 real roots.
Galois theory ( 5 degree polynomial)
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abstract-algebra
1 Answers
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Well, the only options are 1,3 or 5 different real options, as complex roots of real polynomials appear in conjugate pairs and there is at least 1 real root (why?) , so we have to rule the possibility of three real roots and a conjugate pair of complex non-real ones.
But then embedding $\,G\,$ in $\,S_5\,$ we'd get it contains a $\,5-\,$ cycle (as $\,G\,$ has a sbgp. of order 5) but also a transposition (as the complex roots must be mapped to themselves) , and $\,S_5\,$ is generated by any $\,5-\,$ cycle and any transposition...