abstract algebra - Splitting field of $x^3-5 in mathbb{Q}[X]$. Galois group and fields?

I have this multi-part problem I have worked on in Galois Theory. I am particularly unsure abut finding all roots of our polynomial and the action of the Galois group. Also, I cannot see how we can have a subfield of order $4$

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Find the splitting field $L$ of the polynomial $X^3-5 \in \mathbb{Q}[x]$

Let $\mathbb{Q}$ be a field and $f(X) \in \mathbb{Q}[X]$. A field extension $L$ where $\mathbb{Q} \subset L$ is a splitting field for $f(X)$ over $\mathbb{Q}$ if $L=\mathbb{Q}(\alpha_1,..., \alpha_n)$ with $f(X)=c(X-\alpha_1)...(X-\alpha_n)$. $c$ is the leading coefficient of $f(X)$

Here we have $f(X)=X^3-5$ so we must find the roots of $f(X)$. Clearly $\alpha_1=\sqrt[3]{5}$ is a root which means $L=\mathbb{Q}(\sqrt[3]{5})$. Are there any others?

Prove that $Gal(L/\mathbb{Q}) \simeq S_3$ and describe the action of its elements on $L$

The polynomial is a reduced cubic, so the discriminant is given by $-4a^3-27b^2=-27(25)$ which is not square so the Galois group is $S_3$

Is the action given by $\tau: \sqrt[3]{5} \rightarrow -\sqrt[3]{5}$?

Find all subfields of degree $2$ over $\mathbb{Q}$ in $L$

The Galois group has order $6$ so all subgroups will have orders that divide $6$, namely $1, 2, 3$ and $6$.

So the subfield of degree $2$ will correspond to the subgroups of degree $4$ in $S_3$ such as $\{e, (12)\}$

What is the corresponding subfield?

Find all subfields of degree $4$ over $\mathbb{Q}$ in $L$

I am confused here since I thought that degrees of subfields had a 1:1 correspondence with the order of the subgroups. But $S_3$ has no subgroup of order $4$. Where I am going wrong?