Let $p$ be a prime. If $G$ is a normal subgroup of $SL(2, \mathbb Z/p\mathbb Z)$ , then is $G$ also normal in $GL(2, \mathbb Z/p\mathbb Z)$
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For $p\ge5$, $SL(2,p)/\{\pm I\}$ is simple, so the only nontrivial normal subgroup of $SL(2,p)$ is $\{\pm I\}$ which is normal in $GL(2,p)$. You'll then have to treat the cases $p=2$ and $p=3$ separately.
Angina Seng
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Why is $SL(2,p)/{I,-I}$ simple ? – Feb 04 '18 at 07:26
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??? @misao $SL(2,p)$ is not simple. – Angina Seng Feb 04 '18 at 07:27
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I meant to say the quotient ... – Feb 04 '18 at 07:30
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@misao It's $PSL(2,p)$, a well-known example of a simple group of classical Lie type. See textbooks on group theory. – Angina Seng Feb 04 '18 at 07:33
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Can my claim be extended for $\mathbb Z/n$, where $n$ is a product of two distinct primes or a power of prime ? – Feb 04 '18 at 10:12
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Shark the unknown, sir can we replace $2$ by general integer $n$? – MANI Dec 13 '19 at 06:31
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Note the matrix (with $a\neq 0$) $$m_a=\left[ {\begin{array}{cc} a &0\\ 0 &1 \end{array}}\right]$$ has determinant $a$ and inverse $$\left[ {\begin{array}{cc} a^{-1} &0\\ 0 &1 \end{array}}\right]$$ in $GL(2, \mathbb Z/p\mathbb Z)$. Therefore every matrix $g\in GL(2, \mathbb Z/p\mathbb Z)$ can be written as $m_as$ (where $a=\det g$) for some $s\in SL(2, \mathbb Z/p\mathbb Z)$.
Now for all $s\in SL$, $sGs^{-1}=G$. Then for any $g\in GL$, let $g=m_as$, so $gGg^{-1}=m_asGs^{-1}m_a^{-1}=m_aGm_a^{-1}=G$ (somehow)
Can someone help me complete this proof :(
Akababa
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