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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