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11

(Still not quite a complete answer.) An abelian group $A$ does not have a canonically defined torsion-free subgroup in general (the elements of infinite order usually aren't a subgroup). What is canonically defined is a short exact sequence $$0 \to A_T \to A \to A/A_T \to 0$$ where $A/A_T$, which we'll write $A_F$, is the universal torsion-free abelian group ...

2

I know you've gotten one satisfactory answer, but let me weigh in here a bit. I'll first mention that what you present is generally correct, and a valid way to approach this. The idea of "breaking down a (finite) group into smaller pieces" is in fact behind the idea of classifying finite simple groups (groups that cannot be broken down), together ...

2

For example, we can show closure as follows. Let $M_1,M_2$ be the Mobius transformations $$M_1(z) = \frac{a_1 z + b_1}{c_1 z + d_1}, \quad M_2(z) = \frac{a_2 z + b_2}{c_2 z + d_2}.$$ To show that condition 1 holds, show that the function $M_1 \circ M_2$ is a Mobius transformation. $$M_1(M_2(z)) = \frac{a_1 \frac{a_2 z + b_2}{c_2 z + d_2} + b_1}{c_1 \frac{... 2 H^{-1} is the set of all inverses of elements in H i.e. H^{-1}=\{g^{-1}| g\in H\} of course, this notation only makes sense when H\subset G for some group G. Say we take G=\mathbb{Z} and H=\{1,2,3\} then H^{-1}=\{-1,-2,-3\} When H is a group H^{-1} is again all of the inverses of elements in H. So if we again take G=\mathbb{Z}, the ... 1 From what I can tell, the assertion is that the group homomorphism f : \mathbb Z \to \mathbb Z defined by f(r) = kr induces a surjective group homomorphism g(r + m \mathbb Z) = kr + n \mathbb Z if and only if n \,|\, m and \gcd(k, n) = 1. For the case that m = 2^4, n = 2^3, and r = 2, we have that$$k = \frac{rn}{\gcd(m, n)} = \frac{2^4}{2^3} ...

1

Your Iwahori-decomposition computation is a bit too free with quotient computations. It would work fine for vector spaces, which is, in some sense, why you get the correct leading term in your count; but there are additional subtleties on the group level that it does not take into account—among other things, that some of the entries denoted by $\mathcal O$ ...

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