It is known that there is connection between solvability of a group and expressing the roots of a polynomial by radicals, which is something I will study in this semester; however, by the odd order theorem

Every finite group of odd order is solvable.

However, I was wondering is there any relation between the fact that every polynomial of odd order has at least one root, with the fact that every finite group of odd order is solvable ?


Since I haven't studied the main subject in detail (we just started in Ring theory in the graduate Algebra course that I'm taking)

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    $\begingroup$ Just a coincidence. "Odd order" for groups is the number of elements. For polynomials it's the degree. The assertion for polynomials doesn't help find algebraic solutions - it depends on real analysis. $\endgroup$ – Ethan Bolker Dec 3 '18 at 17:19
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    $\begingroup$ Nothing apparent. The statement regarding polynomials is topological, not algebraic. It isn't true over $\mathbb Q$, say. $x^3-2$ has no rational root. It is true over $\mathbb R$ because it goes to $\pm \infty$ as $x$ gets very large or very small. $\endgroup$ – lulu Dec 3 '18 at 17:19
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    $\begingroup$ Not really - equations of odd order having a root is an elementary result in real analysis, whereas the solvability of groups of odd order is a very deep and fiendishly difficult result in abstract algebra. $\endgroup$ – Derek Holt Dec 3 '18 at 17:20
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    $\begingroup$ I'll also add that the Galois groups of odd-order polynomials can very often be even-order. $\endgroup$ – Hempelicious Dec 4 '18 at 0:59
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    $\begingroup$ It's a consequence of continuity. Any continuous function from $\mathbb R \to \mathbb R$ that takes positive and negative values must have a real root. Don't need it to be a polynomial. As I say, the parallel claim is false over $\mathbb Q$. $\endgroup$ – lulu Dec 5 '18 at 18:32

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