Two groups of order 108 The Galois group of the polynomial $x^9+x^3+1$ is a semidirect product of the Heisenberg group $He_3$ and the Klein 4-group V.  But, in the list of groups of order 108 https://people.maths.bris.ac.uk/~matyd/GroupNames/index.html there are only two such groups: one is the semidirect product of $He_3$ with the cyclic group of order 4, and the other is the semidirect product of $He_3$ with $C_4$ acting as
 $C_4/C_2=C_2$.  Which one is my Galois group?
 A: In Maple you can enter
with(GroupTheory):
f := x^9 + x^3 + 1;
G := GaloisGroup(f,x);
Z := Center(G);
Elements(Z);

and you will get the response {()}, indicating that the centre is trivial.  But the second of the two groups that you mention seems to have a central involution, so your Galois group must instead be the first one.  A central involution would correspond to a Galois extension of degree $54$ contained in the full splitting field of $f$.  There is probably a direct field-theoretic argument to show that no such extension exists, without assistance from Maple, but I cannot immediately see one.
UPDATE:
As Jeremy Rouse points out, it seems that the Galois group is not in fact either of the candidates mentioned by the OP.  Instead, we can put $g(y)=y^3+y+1$ so $f(x)=g(x^3)$, and let $K$ and $L$ be the splitting fields of $g$ and $f$.  Each root of $g$ acquires a full set of cube roots in $L$, and the ratio between two such cube roots is a primitive cube root of $1$.  It follows that $L$ contains the field $M=\mathbb{Q}(e^{2\pi i/3})$, which is quadratic over $\mathbb{Q}$.  One can check that $g$ has precisely one real root so complex conjugation gives an involution in $G(K/\mathbb{Q})$ so $G(K/\mathbb{Q})\simeq\Sigma_3$.  It works out that $KM$ is Galois over $\mathbb{Q}$ with $G(KM/\mathbb{Q})\simeq\Sigma_3\times C_2$ (and one can check that this is also the dihedral group of order $12$).  Now let $\alpha$, $\beta$ and $\gamma$ be the roots of $g$.  We can obtain $L$ from $KM$ by adjoining a cube root of $\alpha$ and a cube root of $\beta$ (we then have a cube root of $\gamma$ as well because $\alpha\beta\gamma=-1$, and we have a full set of cube roots because $e^{2\pi i/3}\in M$).  Using this we see that $G(L/KM)=C_3^2$.  We now have an extension 
$$ C_3^2 \to G(L/\mathbb{Q}) \to \Sigma_3\times C_2 $$
Jeremy Rouse's comment suggests that this must be a semidirect product, and Maple agrees with that, but I do not see a direct argument at the moment.
