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

Question

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?

Answer 1

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, $n_{2} = 5$.

Answer 2

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)

Answer 3

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.

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Hope this helps.