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Suppose we have $f \in \mathbb{Q}[X]$, with only real roots. Then the complex conjugation is not an automorphism, but is this enough to say that there exist no order two elements in $\text{Gal}(f)$?

The case I was studying is $f = x^3-4x+2$, I found that it has $3$ real roots. But can I now say that there is no order $2$ element in the Galois group?

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    $\begingroup$ Consider $f(x)=x^2-2$. $\endgroup$ – lulu Jul 7 '18 at 12:03
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    $\begingroup$ For the cubic that you have you can compute the discriminant. It turns out to be $148$, which is not the square of an element of $\mathbb{Q}$. Therefore, the Galois group is $S_3$, which has even order. Therefore, there are order-$2$ elements. $\endgroup$ – user566930 Jul 7 '18 at 12:13
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I don't understand why this question was downvoted. There is a beautiful way to think about such questions when complex conjugation is unavailable, or rather, just by not paying any attention to it at all. Instead of looking at complex conjugation, which is a single little automorphism that may or may not work, reduce the polynomial mod primes. There are lots of primes numbers, and this viewpoint is guaranteed to work.

Theorem: If $f(x) \in \mathbf Z[x]$ is monic irreducible, then the splitting field of $f(x)$ over $\mathbf Q$ has an automorphism of order $2$ if and only if there is some prime $\ell$ such that $f(x) \bmod \ell$ has distinct irreducible factors and their degrees are $1$ and $2$ only.

When I say "degrees are $1$ and $2$ only" I mean the irreducible factors have either degree $1$ or degree $2$, and both options have to occur in the irreducible factorization mod $\ell$.

I can't give a complete proof of this theorem here, as it depends on the (hard) Chebotarev density theorem from algebraic number theory, which incidentally also implies that there will be infinitely many such primes $\ell$ as soon as there is of them.

The meaning of the theorem is illustrated well by examples.

Let's try your cubic polynomial $x^3 - 4x + 2$. It is irreducible either by the Eisenstein criterion or because it is irreducible mod $3$, so let's factor $f(x)$ modulo primes other than $3$ and hope we find one where the irreducible factors have degrees $1$ and $2$.

We have $f(x) \equiv x^3 \bmod 2$, which doesn't help since the irreducible factors are not distinct. But if we look mod $5$ then we're in good shape: $f(x) \equiv (x-4)(x^2+4x+2) \bmod 5$, which has distinct factors and their degrees are $1$ and $2$. Thus, by the theorem, the splitting field of $f(x)$ has an element of order $2$. (More such primes are $13, 17, 19, 23, 29, 31, 47, 59, \ldots$; there is no elementary rule for these primes, but they do continue indefinitely.)

The nice thing about the theorem I stated is that there is a version of it that works for every prime, not just $2$.

Theorem: If $f(x) \in \mathbf Z[x]$ is monic irreducible and $p$ is a prime number, then the splitting field of $f(x)$ over $\mathbf Q$ has an automorphism of order $p$ if and only if there is some prime $\ell$ such that $f(x) \bmod \ell$ has distinct irreducible factors and their degrees are $1$ and $p$ only.

As with the version of this theorem just when $p = 2$, if there is one such prime $\ell$ then there are infinitely many (a positive density's worth).

Let's try a higher degree example: $f(x) = x^8 - 4x^2 + 2$. Again this is irreducible since it is Eisenstein at $2$. If we factor this modulo various primes (with a computer) then we find $f(x) \equiv (x-4)(x-8)(x-11)(x-15)(x^2+11) \bmod 19$, which has degrees $1, 1, 1, 1, 2$, so the Galois group of $f(x)$ over $\mathbf Q$ has an element of order $2$ (but that is easy to see since $\alpha \mapsto -\alpha$ will be a permutation of the roots) and going further we find $$f(x) \equiv (x-11)(x-36)(x^3+13x^2+4x+25)(x^3+34x^2+4x+22) \bmod 47,$$ where the irreducible factors are distinct and the degrees are $1, 1, 3, 3$, so the Galois group of $f(x)$ over $\mathbf Q$ must have an element of order $3$.

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Here is an a;ternative example centred on algebraic extensions (as opposed to polynomials) having even ordered Galois groups.

Consider prime numbers $p$ of the form $p=4k+1$, and let $\zeta = \exp (2\pi i/p)$ a primitive $p$th root of unity. Then the algebraic number $\alpha = \zeta + \bar \zeta$ generates a Galois extension of the rationals having Galois group cyclic of order $(p-1)/2= 2k$, an even number. This field is completely contained in the reals and complex conjugation is not an automorphism (rather is the same as the identity map).

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By Lulu: $f(x) = x^2 - 2$ is a counter example, as we have the automorphism: $\sigma(a+b\sqrt{2}) = a - b\sqrt{2}$.

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