Consider the polynomial $ x^7-13 \in \Bbb{Q}(x)$. Find Galois group and write it as a semi-direct product.


So here is what I have done. I found the dimension of the splitting field over $\Bbb{Q} $ (42, I think) and found the Galois group of splitting field over $\Bbb{Q}[e^{i2\pi/7}]$ (cyclic order 7) and over $Q[ \root7\of{13}] $ (cyclic order 6). I just don't know how to write the whole galois group as a semi-direct product.

  • 2
    $\begingroup$ What have you done so far? And try to write in such a manner that it doesn't look like you're giving orders. $\endgroup$
    – Timbuc
    Sep 15 '14 at 3:08
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    $\begingroup$ Forgive me, I just copied the text. I didn't mean to be rude. $\endgroup$ Sep 15 '14 at 6:49

It sounds like you have ascertained that the splitting field is $F=\Bbb{Q}(\root7\of{13},\zeta_7)$ where $\zeta_7=e^{2\pi i/7}$. Let us denote the key intermediate fields by $L=\Bbb{Q}(\root7\of{13})$ and $K=\Bbb{Q}(\zeta_7)$. We have, indeed, $[F:\Bbb{Q}]=42$, because $[L:\Bbb{Q}]=7$ (Eisenstein), $[K:\Bbb{Q}]=6$ (cyclotomic) and $\gcd(7,6)=1$.

An automorphism $\sigma\in G=Gal(F/\Bbb{Q})$ is fully determined once we know $\sigma(\zeta_7)$ and $\sigma(\root7\of{13})$. There are six possibilities for the former and seven for the latter. Because $|G|=42$ all the combinations occur.

The automorphism $\sigma:\zeta_7\mapsto\zeta_7,\root7\of{13}\mapsto \zeta_7\root7\of{13}$ is clearly of order seven. Equally clearly all the elements of $K$ are fixed by it. Thus we can conclude that $Gal(F/K)=\langle\sigma\rangle\cong C_7$. Because $K/\Bbb{Q}$ is Galois, we know that this copy of $C_7$ is normal in $G$.

The automorphism $c\mapsto c^3$ of $C_7$ is of order six and thus a generator of $Aut(C_7)$. This tells us to also look at the subgroup generated by $\tau:\zeta_7\mapsto \zeta^3, \root7\of{13}\mapsto\root7\of{13}$. It follows easily that this is of order six, and $Gal(F/L)=\langle\tau\rangle\cong C_6$. But as $L/\Bbb{Q}$ is not Galois, this is not a normal subgroup of $G$.

But now you should be well placed to complete the exercise and show that $$ G=Gal(F/K)\rtimes Gal(F/L)\cong C_7\rtimes C_6. $$

Note: The argument is not much different from a proof that the Galois group of $x^3-2$ is $C_3\rtimes C_2\cong S_3$.

For extra credit realize the group $G$ of order 42 as a group of permutations of the roots of $x^7-13$. In other words, give an injective homomorphism $f:G\to S_7$. Clearly $f(\sigma)$ will be the obvious 7-cycle, but do compute $f(\tau)$, and check that conjugating $f(\sigma)$ by $f(\tau)$ gives $f(\sigma^3)$.


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