I'm trying to find a Galois group of $\mathbb Z_2\times \mathbb Z_4$ but, first, I'd like to know if it does exist. I know that a Galois group of order $8$ must be a subgroup of the symmetric group of order $4$, $S_4$. Also, this subgroup must be transitive.

Hence, I should see that $\mathbb Z_2\times \mathbb Z_4$ is not transitive. But how could I do that?

  • $\begingroup$ The subgroups of order 8 in $\,S_4\,$ are the Sylow ones and they all are isomorphic (in fact, conjugate), but they are not what you want but $\,D_4\,$, so not even abelian. If you want to embed $\,C_2\times C_4\,$ in some some symmetric group try with $\,S_6\,$ and the group $\,\langle (12)\,,\,(3456)\rangle\,$ $\endgroup$ – DonAntonio Jan 9 '13 at 12:28

The simplest example is probably the polynomial $X^8+1$, and now I’ll explain why.

It’s a fact that every finite abelian extension of $\mathbb Q$ is contained in a cyclotomic extension $\mathbb Q(\zeta_m)$, $\zeta_m$ being a primitive $m$-th root of unity. The Galois group of $\mathbb Q(\zeta_m)$ over $\mathbb Q$ is the group of units of the ring $\mathbb Z/m\mathbb Z$. With this information, you can get any finite abelian group as the quotient of a $(\mathbb Z/m\mathbb Z)^\times$, and so an extension of $\mathbb Q$ with that Galois group.

Now, $\mathbb Z/16\mathbb Z$ has for its units the odd numbers modulo $16$, as a multiplicative group, and you easily check that this has the shape $\mathbb Z/2\mathbb Z\times\mathbb Z/4\mathbb Z$, generators being $-1$ and $5$. The polynomial for the primitive sixteenth roots of unity is $(X^{16}-1)/(X^8-1)=X^8+1$, and there you have it.

  • $\begingroup$ This is the answer that I like the most. It really helped me. I have a question which is so closely related that I don't think necessary to open a new thread. $X^8-1$ is not irreducible over $\mathbb Q$. Should there be an irreducible polynomial which splitting field is also isomporphic to this multiplicative group? $\endgroup$ – synack Jan 9 '13 at 17:50
  • $\begingroup$ In speaking of $X^8-1$, you’re asking about the eighth roots of unity. A primitive such is $(1+i)/\sqrt2$, and its minimal polynomial is $X^4+1$. $\endgroup$ – Lubin Jan 10 '13 at 16:21

Consider the splitting field of $x^5-1$, $x^2-2$ on $\mathbb{Q}$.


A) The Galois group of a degree $n$ polynomial need not be a transitive subgroup of $S_n$. E.g. what is the Galois group of a biquadratic extension? What you have in mind is that the Galois group of an irreducible polynomial is a transitive subgroup of $S_n$, where $n$ is the degree of the polynomial.

B) Why does a Galois group of order 8 have to be a subgroup of $S_4$? What tells you that you cannot have, say, a cyclic group of order 8 as a Galois group (which would then be a transitive subgroup of $S_8$)?

So with these remarks in mind, perhaps you should ask the precise question that you meant to ask.

Anticipating what your question might be: MAGMA tells me that the group generated by $$ (1, 4)(2, 5)(3, 8)(6, 7),\\ (1, 5, 6, 8)(2, 7, 3, 4),\\ \text{and } (1, 6)(2, 3)(4, 7)(5, 8) $$ is a transitive subgroup of $S_8$ that is isomorphic to $C_2\times C_4$. It also tells me that there are no such subgroups in any $S_n$ for $n$ smaller than 8.

  • $\begingroup$ the group I wrote in my comment above is $\,C_2\times C_4\,$ and embedded in $\,S_6\,$ ... $\endgroup$ – DonAntonio Jan 9 '13 at 13:35
  • $\begingroup$ It is not a transitive subgroup of $S_6$. I am only guessing what the OP actually wants. $\endgroup$ – Alex B. Jan 9 '13 at 13:44
  • $\begingroup$ Ok, but the OP was wrong both in thinking $\,C_2\times C_4\,$ can be embedded in $\,S_4\,$ and, even worse, in thinking it must be transitive there. I'm not sure now whether he only wants to embed his group in some permutation group or if he also wants it to be transitive...I think the last property may not be that important to him, but if it is yours is the correct example. $\endgroup$ – DonAntonio Jan 9 '13 at 13:51
  • $\begingroup$ Thanks for your answer but my question is about the identification of $C_2\times C_4$ as a Galois group of order $8$. My mistake was to claim transitivity because I was looking for an irreducible polynomial which splitting field defined that Galois group. $\endgroup$ – synack Jan 10 '13 at 10:53

Actually, the simplest example is $\Bbb Q(\zeta_{15})$, since cyclotomic extensions are well-known.

A quick calculation shows $\operatorname{Gal}(\Bbb Q(\zeta_{15})/\Bbb Q) = (\Bbb Z/15\Bbb Z)^\times = (\Bbb Z/3\Bbb Z)^\times \times (\Bbb Z/5\Bbb Z)^\times = C_2 \times C_4$.


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