# Galois Theory Problem

Let $L$ be a Galois extension of fields such that $Gal(L/K)=GL_{2}(\mathbb{F}_{p})$. Let $L_{1}$ and $L_{2}$ be subfields of $L$ containing $K$ corresponding to subgroups $H_{1}=SL_{2}(\mathbb{F}_{p})$ and $H_{2}$ be the subgroup of upper triangular 2x2 matrices with the diagonal entries from $\mathbb{F}_{p}^{\times}$ and the remaining entry from $\mathbb{F}_{p}$.

Show that $L_{1}L_{2}$ is not a Galois extension of $K$ and compute $Gal(L_{1}L_{2}:L_{2})$.

I have computed that the degree of $[L_{1}L_{2}:K]$ is $p^{2}-1$. The subgroup corresponding to $L_{1}L_{2}$ is the subgroup $H_{1}\cap H_{2}$. To see that $L_{1}L_{2}$ is not Galois over $K$, it amounts to showing that $H_{1}\cap H_{2}$ is not normal in $GL_{2}(\mathbb{F}_{p})$. I am wondering is there a better method to do this rather just working out mechnically.

• Have you tried conjugating with $$\pmatrix{0&1\\1&0}?$$ Sep 7 '11 at 5:56

The intersection $H_1\cap H_2$ contains $\begin{pmatrix}1&1\\\\0&1\end{pmatrix}$ which is conjugate to $\begin{pmatrix}1&0\\\\1&1\end{pmatrix}$ which is not in $H_1\cap H_2$.
It follows that $H_1\cap H_2$ is not normal.
As Jyrki Lahtonen's comment indicates, the actual check is very easy. Since it's clear how to calculate the determinants of upper triangular matrices, it's also clear what the intersection of $H_1$ and $H_2$ is. After that, just conjugate by something suitable.
But if you are willing to use some more sophisticated group theory, then it becomes even easier. Namely, $H_1\cap H_2$ properly contains the scalar matrices in $\text{SL}_2$, which constitute a normal subgroup, $Z$, say (because this is in fact the centre of $\text{SL}_2(\mathbb{F}_p)$). Thus $H_1\cap H_2$ is the lift of some non-trivial subgroup of $\text{SL}_2(\mathbb{F}_p)/Z$, and by one of the isomorphism theorems it is normal in $\text{SL}_2(\mathbb{F}_p)$ if and only if its image in $\text{SL}_2(\mathbb{F}_p)/Z$ is normal. But $\text{SL}_2(\mathbb{F}_p)/Z=\text{PSL}_2(\mathbb{F}_p)$ is well known to be simple, so you are done.