International Mathematics Competition
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IMC2021: Day 2, Problem 6

Problem 6. For a prime number \(\displaystyle p\), let \(\displaystyle \mathrm{GL}_2(\ZZ/p\ZZ)\) be the group of invertible \(\displaystyle 2 \times 2\) matrices of residues modulo \(\displaystyle p\), and let \(\displaystyle S_p\) be the symmetric group (the group of all permutations) on \(\displaystyle p\) elements. Show that there is no injective group homomorphism \(\displaystyle \varphi : \mathrm{GL}_2(\ZZ/p\ZZ) \to S_p\).

Thiago Landim, Sorbonne University, Paris

Solution. For \(\displaystyle p=2\), just note that \(\displaystyle \mathrm{GL}_2(\ZZ/2\ZZ)\) has more than \(\displaystyle 2=\vert S_2\vert\) elements.

From now on, let \(\displaystyle p\) be an odd prime and suppose that there exists such a homomorphism.

The matrix

\(\displaystyle A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \)

has order \(\displaystyle p\) and commutes with the matrix

\(\displaystyle B = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \)

of order \(\displaystyle 2\), hence \(\displaystyle AB\) has order \(\displaystyle 2p\). But there is no permutation in \(\displaystyle S_p\) of order \(\displaystyle 2p\) since only \(\displaystyle p\)-cycles have order divisible by \(\displaystyle p\), and their order is exactly \(\displaystyle p\).

IMC
2021

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