# The galois group of polynomial ($x^5-3$) over $\mathbb{Q}$ [duplicate]

Let the splitting field of $$x^5-3$$ over $$\mathbb{Q}$$ be $$L$$.

I solved the 3 statement,

1. $$L = Q(\sqrt[5]{3}, w) (w=e^{2\pi i/5})$$
2. $$\operatorname{Gal}(L/\mathbb Q)$$ has unique normal subgroup $$H$$ of which the order is $$5$$, corresponding to $$\mathbb Q(w)$$ by Galois theory.
3. the fixed field of $$H$$ is $$\mathbb Q(w)$$

Then, I hit the following:

1. Distinguish $$\operatorname{Gal}(L/\mathbb Q)$$ is abelian or not
2. The number of subgroups of $$\operatorname{Gal}(L/\mathbb Q)$$ of which order is $$4$$.

How can I solve this?

• If Gal$(L/Bbb Q)$ were abelian, then every subgroup would be normal; is that the case for this group? Commented May 2, 2020 at 6:59
• See also this post or this post. Commented May 2, 2020 at 8:35

Below is an approach using elementary calculations.

We first find the number of subgroups of order $$4$$.

Denote $$\operatorname{Gal}(L/\mathbb Q)=G$$. Note that the order of $$G$$ is $$|H|\times|\operatorname{Gal}(L^H/\mathbb Q)|=20$$, and every element in a subgroup of order $$4$$ has order dividing $$4$$. So we count the number of elements of order dividing $$4$$.

Let $$\sigma\in G$$ be an element of order dividing $$4$$. Since $$\sigma$$ is an automorphism, it must send $$w$$ to one of $$w^i,\,i=1,\ldots,4$$. If $$\sigma(w)=w$$, then its action is completely determined by $$\sigma(\sqrt[5]3)=\sqrt[5]3w^k$$ for some $$k=0,\ldots,4$$. So $$\sigma$$ has order dividing $$5$$, a contradiction unless $$\sigma=1_G$$. Thus suppose $$\sigma$$ sends $$w$$ to $$w^i,\,i=2,3,4$$.

Suppose $$\sigma(w)=w^i$$ for some $$i=2,3,4$$, and $$\sigma(\sqrt[5]3)=\sqrt[5]3\cdot w^k$$, for some $$k=0\ldots,4$$. It follows that $$\sigma^4(w)=w$$, and \begin{align} \sigma^2(\sqrt[5]3)&=\sqrt[5]3\cdot w^{k+ik}\\ \sigma^3(\sqrt[5]3)&=\sqrt[5]3\cdot w^{k+i(i+1)k}\\ \sigma^4(\sqrt[5]3)&=\sqrt[5]3\cdot w^{k+ik+i^2(i+1)k}=\sqrt[5]3\cdot w^{k(i+1)(i^2+1)}. \end{align} It is easy to check that for $$i=2,3,4$$, we have $$5\mid(i+1)(i^2+1)$$, so $$\sigma^4(\sqrt[5]3)=\sqrt[5]3$$.

Consequently, an element $$\sigma\in G$$, which is not equal to $$1_G$$, is of order dividing $$4$$ if and only if it sends $$w$$ to $$w^i$$ for some $$i=2,3,4$$ and sends $$\sqrt[5]3$$ to $$\sqrt[5]3\cdot w^k$$ for some $$k=0,\ldots,4$$.

We shall now decide how many subgroups are formed by these elements. Denote the automorphism sending $$w$$ to $$w^i$$ and $$\sqrt[5]3$$ to $$\sqrt[5]3\cdot w^k$$ as $$\sigma_{i,k}$$.

Observe that the above calculation shows that $$\sigma_{i,k}^2=\sigma_{i^2,(i+1)k}$$. So $$\sigma^2=1_G$$ if and only if $$i=4$$. This also shows that if $$i,j\ne4$$ and $$\sigma_{i,k}^2=\sigma_{j,\ell}^2$$, then $$i^2\equiv j^2\pmod5$$ and $$(i+1)k\equiv(j+1)\ell\pmod5$$. Then $$i^2\equiv j^2\pmod5$$ implies $$i\equiv\pm j\pmod5$$. And $$\ell\pmod5$$ is uniquely determined by the congruence equation $$(i+1)k\equiv(j+1)\ell\pmod5$$. So, for any $$i=2,3$$, and $$k=0,\ldots,4$$, there are exactly two $$\sigma_{j,\ell}$$ such that $$\sigma_{i,k}^2=\sigma_{j,\ell}^2$$. Since $$\sigma_{i, k}$$ is of order $$4$$, these two elements are exactly $$\sigma_{i, k}$$ and $$\sigma_{i, k}^3$$.

Thus we can divide the set of elements $$\ne1_G$$ of order dividing $$4$$ into $$5$$ mutually disjoint sets of the form $$\left\{\sigma_{i, k},\sigma_{i,k}^2,\sigma_{i,k}^3\right\}$$. We see immediately that they form (with $$1_G$$ adjoined) $$5$$ subgroups of $$G$$ of order $$4$$.

Finally, by an elementary calculation, we see that $$\sigma_{i, k}\cdot\sigma_{j,\ell}=\sigma_{ij,k+i\ell}.$$ Clearly this is non-abelian.

Hope this helps.

As cited in the comments, if Gal$$(\mathbb{Q}(\zeta_{5},\sqrt[5]3)/\mathbb{Q})$$ would be abelian every subextension would be normal, but $$\mathbb{Q}(\sqrt[5]3)/\mathbb{Q}$$ isn't, do you see why ?

As the matter of the the number of subgroups of order $$4$$ since the group has order $$20$$ we note that the subgroups of order $$4$$ coincide with the $$2-$$Sylow of $$G$$ = Gal$$(\mathbb{Q}(\zeta_{5},\sqrt[5]3)/\mathbb{Q})$$,

From Sylow theory we know the number $$n_{2}$$ of $$2$$-Sylow is such that $$n_{2} \equiv 1$$ mod($$p$$) and $$n_{2} \mid 5$$, of course since the group is not abelian the number of $$2$$-Sylows can't be 1 (Why ?)

So we have that the number we were looking for, $$n_{2} = 5$$.

If G(L/Q) abelian group. Then every sub group of G(=G(L/Q)) is normal so every sub field of L is splitting field over Q but we have one counter example Q(3^1/5)