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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.

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Have you tried conjugating with $$\pmatrix{0&1\\1&0}?$$ – Jyrki Lahtonen 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.

Everyone should prove at some time in their lives that every matrix is conjugated to its transpose, in fact :)

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Please take a look at: – George Sep 7 '11 at 13:55

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.

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You've written a great candidate for :) – Mariano Suárez-Alvarez Sep 7 '11 at 7:12

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