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Nov 18, 2018 at 14:31 history edited Mikhail Borovoi CC BY-SA 4.0
Mistake corrected following comments of Arkandias.
Nov 16, 2018 at 20:00 comment added Mikhail Borovoi @Arkandias: The maps $G(k)\to\mathbb{Z}_2/2\mathbb{Z}_2$ that I constructed is not a homomorphism! I will edit my answer tomorrow.
Nov 16, 2018 at 19:47 comment added Arkandias Finally, this subgroup has cardinal $(q^n-1)/(q-1)$ where $n^2$ is the dimension of $A$ over $k$. So in your case ($k=\mathbb{Q}_2$ and $A$ is the quaternion algebra), the order of $G_0/H_1$ should be $3$ according to this corollary, so in particular there should not be any morphism into $\mathbb{Z}/2\mathbb{Z}$. Yet, I do not see anything wrong with your construction...
Nov 16, 2018 at 19:42 comment added Arkandias $k$ is a $p$-adic field with ring of integers $\mathfrak{o}$ and maximal ideal $\mathfrak{p}$, $A$ is a central division algebra over $k$ with ring of integers $\mathfrak{D}$ and maximal ideal $\mathfrak{P}$. The corollary claims that the commutator of $G_0 := \mathrm{SL}(1,A)$ is the congruence subgroup $H_1$ consisting of those elements $g \in G_0$ which are $\equiv 1 \bmod \mathfrak{P}$. So the abelianization is isomorphic to $G_0/H_1$, which in turn is isomorphic to the norm $1$ subgroup of the field extension $\mathfrak{D}/\mathfrak{P}$ of $\mathfrak{o}/\mathfrak{p}$.
Nov 16, 2018 at 13:45 comment added Mikhail Borovoi @Arkandias: Do I understand correctly that in our case $q=2$, so $q+1=3$ ?
Nov 16, 2018 at 13:44 comment added Mikhail Borovoi @Arkandias: Could you please explain how the corollary to Theorem 21 implies that the abelianization of ${\rm SL}(1,D)$ has order $q+1$?
Nov 15, 2018 at 21:36 comment added Arkandias Yes, but the Corollary includes that case (see just above the statement, and most of the proof is about the dyadic quaternion algebra case).
Nov 15, 2018 at 18:58 comment added Mikhail Borovoi @Arkandias: In Theorem 21 Riehm assumes that $D$ is not a dyadic quaternion algebra.
Nov 15, 2018 at 16:02 comment added Arkandias This example seems to contradict the Corollary to Theorem 21 in Riehm "The norm 1 group of a $p$-adic division algebra", which implies that the abelianization of $\mathrm{SL}(1,D)$ has order $q+1$. Am I missing something?
May 20, 2016 at 9:29 vote accept Daniel Loughran
May 20, 2016 at 9:27 history edited Mikhail Borovoi CC BY-SA 3.0
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May 20, 2016 at 5:19 history edited Mikhail Borovoi CC BY-SA 3.0
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May 19, 2016 at 20:19 history edited Mikhail Borovoi CC BY-SA 3.0
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May 19, 2016 at 20:06 history edited Mikhail Borovoi CC BY-SA 3.0
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May 19, 2016 at 19:50 history edited Mikhail Borovoi CC BY-SA 3.0
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May 19, 2016 at 19:44 history answered Mikhail Borovoi CC BY-SA 3.0