Condition for having order 2 elements in Galois group for polynomials over $\mathbb{Q}$ 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?

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).
