# Galois Groups are isomorphic to subgroups of symmetric groups.

I am currently working through Joseph Rotman's book "Galois Theory" and am trying to prove the following theorem.

If $$f(x) \in F[x]$$ has $$n$$ distinct roots in its splitting field $$E$$, then Gal($$E/F$$) is isomorphic to a subgroup of the symmetric group $$S_n$$, thus its order is a divisor of $$n!$$.

I have struggled to understand the proof in the book and have therefore tried to write my own proof of this fact, I have found some questions close to this on this site, but haven't been able to find exactly what I'm looking for.

My proof (so far) is as follows:

Proof:

Let $$X = \{\alpha_1, ... , \alpha_n \}$$ be the set of roots of $$f(x)$$, as $$E$$ is the splitting field of $$f$$, it may be written as $$E = F(\alpha_1, ... , \alpha_n)$$ and noting that $$\alpha_i \neq \alpha_j$$ for $$i\neq j$$. Let $$\sigma \in$$ Gal($$E/F$$) then $$\sigma(X) = X$$ (a result from an earlier lemma). Now I would like to construct an injective homomorphism from $$Gal(E/F)$$ to $$S_n$$, my guess is the map which sends $$\sigma$$ to its restriction in $$X$$, which I shall denote by $$\sigma_X$$. Let $$g$$ be this map, then it is clear that $$g$$ is a homomorphism.

Now I try to show that $$g$$ is injective, consider $$\sigma , \phi \in Gal(E/F)$$ where $$\sigma \neq \phi$$, both of these functions are F-automorphisms of $$E/F$$, so they "fix" $$F$$, then we can conclude that $$\sigma$$ and $$\phi$$ can be different if and only if they act differently on $$X$$, which in the language of mathematics is precisely $$\sigma \neq \phi \Leftrightarrow \sigma_X \neq \phi_X$$. This is the definition of an injective function.

This means that $$|Gal(E/F)| \leq |S_n| = n!$$, however due to having constructed a homomorphism, we have mapped a group to another group, hence we have mapped $$Gal(E/F)$$ to a subgroup of $$S_n$$, whose order divides $$|S_n|$$ and therefore $$|Gal(E/F)|$$ divides $$n!$$, completing the proof.

Question

I am wondering whether my proof for this is correct, I think that it is mostly there, with the statement

we can conclude that $$\sigma$$ and $$\phi$$ can be different if and only if they act differently on X

perhaps being the only issue.

• If $\sigma$ fixes all elements of $X$, then it fixes the splitting field $E$ as well. So $\sigma$ must be the identity and hence your map is injective. – hellHound Dec 30 '18 at 17:26
• Use $\operatorname{Gal}(E/F)$ for $\operatorname{Gal}(E/F)$. – Shaun Dec 30 '18 at 17:59
• > implying $E/F$ is galois – Kenny Lau Dec 31 '18 at 13:54

Assume $$\sigma_X = \varphi_X$$, i.e. $$\forall i, \sigma(\alpha_i) = \varphi(\alpha_i)$$. We shall show that $$\sigma = \varphi$$, i.e. $$\forall x \in E, \sigma(x) = \varphi(x)$$.
Now note that $$E = F(\alpha_1, \cdots, \alpha_n)$$, and every $$\alpha_i$$ is algebraic over $$F$$, so every $$x \in E$$ can be expressed as a polynomial in $$(\alpha_i)_{i=1}^n$$ with coefficients in $$F$$, say $$x = \sum f_j \alpha_{1}^{v_{j1}} \alpha_{2}^{v_{j2}} \cdots \alpha_{n}^{v_{jn}}$$. Then:
$$\begin{array}{rcll} \sigma(x) &=& \sigma \left( \sum f_j \alpha_{1}^{v_{j1}} \alpha_{2}^{v_{j2}} \cdots \alpha_{n}^{v_{jn}} \right) \\ &=& \sum \sigma \left( f_j \alpha_{1}^{v_{j1}} \alpha_{2}^{v_{j2}} \cdots \alpha_{n}^{v_{jn}} \right) & \text {\sigma preserves addition} \\ &=& \sum \sigma \left( f_j \right) \sigma \left( \alpha_{1} \right)^{v_{j1}} \sigma \left( \alpha_{2} \right)^{v_{j2}} \cdots \sigma \left( \alpha_{n} \right)^{v_{jn}} & \text {\sigma preserves multiplication} \\ &=& \sum f_j \sigma \left( \alpha_{1} \right)^{v_{j1}} \sigma \left( \alpha_{2} \right)^{v_{j2}} \cdots \sigma \left( \alpha_{n} \right)^{v_{jn}} & \text {\sigma fixes F} \\ &=& \sum f_j \varphi \left( \alpha_{1} \right)^{v_{j1}} \varphi \left( \alpha_{2} \right)^{v_{j2}} \cdots \varphi \left( \alpha_{n} \right)^{v_{jn}} & \forall i, \sigma(\alpha_i) = \varphi(\alpha_i) \\ &=& \sum \varphi \left( f_j \right) \varphi \left( \alpha_{1} \right)^{v_{j1}} \varphi \left( \alpha_{2} \right)^{v_{j2}} \cdots \varphi \left( \alpha_{n} \right)^{v_{jn}} & \text {\varphi fixes F} \\ &=& \sum \varphi \left( f_j \alpha_{1}^{v_{j1}} \alpha_{2}^{v_{j2}} \cdots \alpha_{n}^{v_{jn}} \right) & \text {\varphi preserves multiplication} \\ &=& \varphi \left( \sum f_j \alpha_{1}^{v_{j1}} \alpha_{2}^{v_{j2}} \cdots \alpha_{n}^{v_{jn}} \right) & \text {\varphi preserves addition} \\ &=& \varphi(x) \end{array}$$