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I am trying to see that $\Gamma_2$, defined as the kernel of the natural surjective map $\text{GL}_2(\mathbb Z_p)\to \text{GL}(\mathbb F_p)$ is a pro-$p$ group. So I'm trying to show that every finite quotient is a $p$-group.

As my first attempt, I'm trying to show that if there is a map $$\phi: \Gamma \to \mathbb Z/l\mathbb Z$$ then it has to be trivial. Since the right side is abelian, the kernel of this map contains the commutator subgroup.

I'm guessing that I want to show that the commutator is open, therefore closed and of finite index so that the left side will be recognizable $p$-group but beyond that I don't know. Even this will not solve the problem because I need to replace the $\mathbb Z/l\mathbb Z$ above with an arbitrary simple group and that case seems hard..

EDIT: I just noticed that $\Gamma_2$ contains the subgroup of matrices of the form $\left[\begin{array}{cc} 1 & m \\ 0 & 1 \\ \end{array}\right]$ for all $m\in p\mathbb Z_p$ so maybe this is helpful?

Thanks for your help and I hope I've shown enough of my work. If not please let me know and I'll try to add a bit more.

Ariadne
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1 Answers1

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For any whole number $n$ there is a surjection $\mathrm{GL}_2(\mathbb{Z}/p^n\mathbb{Z})\to\mathrm{GL}_2(\mathbb{Z}/p\mathbb{Z})$, Let $K_n$ be the kernel of this map. Then we have an infinite commutative diagram

$$\require{AMScd} \begin{CD} & & \vdots \\ & K_3 @>>> \mathrm{GL}_2(\mathbb{Z}/p^3\mathbb{Z}) \\ @VVV @VVV \\ & K_2 @>>> \mathrm{GL}_2(\mathbb{Z}/p^2\mathbb{Z}) \\ @VVV @VVV \\ & K_1 @>>> \mathrm{GL}_2(\mathbb{Z}/p\mathbb{Z}) \end{CD}$$

The kernel of $\mathrm{GL}_2(\mathbb{Z}_p)\to\mathrm{GL}_2(\mathbb{Z}/p\mathbb{Z})$ will be the inverse limit $\varprojlim K_n$.

Now compute $|K_n|$ to find it is a $p$-group for all $n$. Note all matrices in $K_n$ are expressible in the form $I+pA$ where $I$ is the identity matrix and $A$ is an arbitrary $2\times 2$ matrix over $\mathbb{Z}/p^n\mathbb{Z}$.

anon
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  • Thank you. Just to confirm: $K_n$ is basically $(\mathbb Z/p^{n-1}\mathbb Z)^4$ so it's a $p$-group right? – Ariadne Nov 27 '16 at 04:49
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    @Ariadne I wouldn't call it that, but there's an obvious bijection with the entries yes. – anon Nov 27 '16 at 04:51