# Is there a non trivial normal subgroup of a group $G$, where $|G|=pm, \ \gcd(p,m)=1$?

In fact, the exercise I'm in is this:

Suppose you have an irreducible polynomial $f(x)\in \mathbb{Q}[x]$ of degree $p$, where $p$ is a prime. Also suppose that $K$ is a splitting field of $f(x)$ over $\mathbb{Q}$.

Prove that:

1. $[K:\mathbb{Q}]=pm$, where $\gcd(p,m)=1$
2. If $H$ is a normal subgroup of $\mathrm{Gal}(K,\mathbb{Q})$ of order $|H|=m$ then $m=1$.

The first part was easy but the second part has something to do with Sylow theorems in which I'm not very keen on.

• $K$ is called a $\textit{splitting field}$ of $f(x)$. – Ken Duna Jun 16 '16 at 1:23

Answer to your second question: Since $$H$$ is normal in $$G=Gal(K/\mathbb{Q})$$, $$H$$ corresponds to a field $$E$$ such that $$\mathbb{Q} \subseteq E \subseteq K$$ (more preciously $$E=K^{H}$$). Now $$H$$ being normal in $$G$$ we must have that $$E$$ is a normal extension of $$\mathbb{Q}$$. (Because $$\bigcap_{g\in G} gHg^{-1}$$ corresponds to $$\prod_{\sigma \in G}\sigma(E)$$ and $$H$$ being normal in $$G$$ we've $$\prod_{\sigma \in G}\sigma(E)$$=E, which forces $$E/\mathbb{Q}$$ is anormal extension). And as $$K/\mathbb{Q}$$ is itself separable (as it is a Galois extension), $$E/\mathbb{Q}$$ is also separable and hence Galois of degree $$p$$.

So now by Primitive Element theorem there exists $$b\in E$$ such that $$E=\mathbb{Q}(b)$$. Now if $$b$$ is root of $$f(x)$$ then since $$E/\mathbb{Q}$$ is normal $$E$$ would become a splitting field of $$f(x)$$ forcing $$m=1$$. So WLOG $$b$$ is not a root of $$f(x)$$. Now pick a root of $$f(x)$$ $$a\in F$$ to and consider $$\mathbb{Q}(a,b)/\mathbb{Q}$$.

Note that $$[\mathbb{Q}(b):\mathbb{Q}]=p$$ so now if we can show $$[\mathbb{Q}(a,b):\mathbb{Q(b)}]=p$$ then we will get $$p^2 \mid [K:\mathbb{Q}]$$ which is a contradiction from first part. So, now we're just left to prove $$[\mathbb{Q}(a,b):\mathbb{Q(b)}]=p$$

Suppose not, which means $$f(x)$$ is reducible over $$\mathbb{Q}(b)$$. Now note that $$f(x)$$ is irreducible over $$\mathbb{Q}$$ and $$\mathbb{char}(\mathbb{Q})=0$$so $$f(x)$$ is seperable, so it has no multiple roots in any extension of $$\mathbb{Q}$$ and so, $$f$$ must split into linear factors over $$\mathbb{Q}(b)[x]$$ (if not then its has a factor of form $$g(x)^r$$ with $$r>1$$ and $$g$$ is irreducible over $$\mathbb{Q}(b)[x]$$, but then any root of $$g$$ in some extension comes as a root of $$f$$ with multiplicity more than $$1$$, contradiction) and so $$b$$ is a root of $$f$$ which is a contradiction to our assumption that $$b$$ is not a root of $$f$$ and so we're done.

There is the simple counterexample of $S_3$. Its order is $6=3\times2$ but its order 2 subgroups are not normal.

• In fact, this is exactly what I' m trying to prove, so that' s makes it an example instead of a counter one. – richarddedekind Jun 16 '16 at 20:05

To answer to the question of the title. Consider the group $G=\mathbb{Z}/6\mathbb{Z}$, we have $\mid G \mid = 3\times 2$ and $\gcd(2,3)=1$. Since $G$ is abelian then $H=\langle 2 \rangle \simeq \mathbb{Z}/3\mathbb{Z}$ is a non trivial normal subgroup of $G$.

• This does not answer the question. In fact the statement is not true in general. Consider $S_3$. Its order is $6=3\times 2$. But each subgroup of order 2 is not normal. – Ken Duna Jun 16 '16 at 13:15
• If $G$ is a Galois group of order 6 isomorphic to $\mathbb{Z}_6$ you are right and it is a valid counterexample. But, if there are no splitting fields with Galois group isomorphic to $\mathbb{Z}_6$? In fact, I tried to find one but I always end up into the symmetric group $S_3$. – richarddedekind Jun 17 '16 at 3:40