We reprint an old math SE question here (see <https://math.stackexchange.com/questions/1241224/composition-of-polynomials-and-galois-theory>):



> 
> "
> Let $f(x)$ be a polynomial of degree $n$ over $\mathbb{Q}$, with Galois group
> isomorphic to the symmetric group $S\_n$. How do I show that $f$ cannot be expressed
> as a composition $g(h(x))$ of two polynomials $g$ and $h$ of degrees $> 1$.
> "
> 
> 
> 


This old question does not have an answer, but one comment refers to the article


<http://www.ccms.or.kr/data/pdfpaper/jcms22_3/22_3_497.pdf>


of Choi. Therein, in the paragraph after Lemma 3.2, it is written:



> 
> "One of the important results about Gal$(f(g(x))/K)$ is that the Galois group is a
> wreath product of certain groups ([6])."
> 
> 
> 


Here, $K$ denotes any field and the reference $[6]$ points to the article


"The Galois theory of iterates and composites of polynomials" by Odoni.


Alas, Choi does not give a particular Lemma or Theorem of $[6]$ as a reference.


The closest we could find is Lemma $4.1$ in $[6]$:


$K$ is an arbitrary field.



> 
> LEMMA $4.1.$ Let $f(g(X))$ be separable
> over $K$, and let deg$(f)= k$, deg$(g)=l$,
> with $k,l\geq 1$. Then $f(X)$ is also
> separable over $K$. Let $\mathcal{F}$ be
> Gal $f(X)/K$, identified with a subgroup of
> the permutations of its zeros in the usual
> way. Then there is an **injective**
> homomorphism of Gal $f(g(X))/K$ into the
> wreath product of $\mathcal{F}$ with the
> symmetric group $S\_l$.
> 
> 
> 


The question is now:



> 
> How to derive the statement in Choi's article from Lemma $4.1$ of Odoni's article?
> 
> 
> 


Or is there another result of Odoni's article needed?


Any additional references are very welcome.


EDIT:


I am interested in the question, if the Galois group of two such polynomials is a wreath product in a non-trivial way.


Thank you very much for the help.



On the Galois group of the compositions of polynomials
Since $\mathcal{F} \le S\_k$, the wreath product of $\mathcal{F}$ with $S\_l$ is no larger than the wreath product of $S\_k$ with $S\_l$. This has cardinality
$$ (l!)^k k! < (k\cdot l)! = |S\_{k\cdot l}|.$$
(Perhaps the wreath product we need is the opposite one, with cardinality $(k!)^l l!$. The result is the same.)


To see the inequality, write
$$\frac{(k\cdot l)!}{(l!)^k} = \frac{(1\cdot\ldots\cdot l)}{(1\cdot\ldots\cdot l)}\cdot\frac{ ((l+1)\cdot\ldots\cdot (2l))}{(1\cdot\ldots\cdot l)}\cdot\ldots\cdot \frac{((kl-l+1)\ldots\cdot(kl))}{(1\cdot\ldots\cdot l)}$$
$$> \frac{l}{l}\cdot\frac{2l}{l}\cdot\ldots\cdot\frac{kl}{l} =k!,$$
where to pass from the first line to the second we take the last factor from the numerator and denominator of each fraction (it is $>$ and not just $\ge$ if $k>1$).


Therefore $\mathrm{Gal}(f\circ g(X)/K)$ has cardinality less than $S\_{k\cdot l}$.




---


Edit:
Regarding whether the Galois group is the wreath product, the answer is no. Consider the Galois group of $f\circ f(x)$ over $\mathbb{Q}$, where $f(x)=x^2 - 2$. Its order is $4$, while the wreath product of $S\_2$ with itself has order $8$.


