As you remark the part in bold is not completely correct. For it to be it should be that the Galois group acts transitively on
the four roots, and then it'd follow that the group is isomorphic to a subgroup of .
There's another way to show that : first, it's easy to see that cannot be abelian and that
its order is divisible by four and divides 8, so it is either . Nevertheless, this extension has a non-normal
subextension, , from where it follows that there is a non-normal subgroup of
of index 4, and since all the subgroups are normal then we're done.