English (United Kingdom)

https://curated-unify.zendy.io/wp-json/zendy-region/v1/featured_content/oa?rat=en

https://curated-unify.zendy.io/wp-json/zendy-region/v1/highlighted_journal/

Global: Zendy Plus annual

Presents the access of premium content as premium feature

Premium Content

Presents the keyphrase highlighting as premium feature

Keyphrase Highlighting

Presents the summarisation as premium feature

Summarisation

Insights

Presents the pdf analysis as premium feature

PDF Analysis

Presents the zaia usage as premium feature

ZAIA

Global: Zendy Plus monthly

Global: Zendy Tools annual

Global: Zendy Tools monthly

Global: Zendy Free Plan

For all $m \geq 1$, we prove that the abelianization of$\operatorname{SL}_2(\mathbb{Z}[\frac{1}{m}])$ is (1) trivial if $6 \mid m$;(2) $\mathbb{Z} / 3\mathbb{Z}$ if $2 \mid m$ and $\gcd(3,m)=1$; (3) $\mathbb{Z}/ 4 \mathbb{Z}$ if $3 \mid m$ and $\gcd(2,m)=1$; and (4) $\mathbb{Z} /{12}\mathbb{Z} \cong \mathbb{Z} / 3\mathbb{Z} \times \mathbb{Z} / 4\mathbb{Z}$if $\gcd(6,m)=1$. This completes known computational results of Bui Anh & Ellisfor $m \leq 50$. The proof is completely elementary, and in particular does notuse the congruence subgroup property. We also find a new presentation for$\operatorname{SL}_2(\mathbb{Z}[\frac{1}{2}])$. This presentation has twogenerators and three relators. Thus,$\operatorname{SL}_2(\mathbb{Z}[\frac{1}{2}])$ admits a presentation withdeficiency equal to the rank of its Schur multiplier. This also gives new andvery simple presentations for the finite groups $\operatorname{SL}_2(\mathbb{Z}/ m \mathbb{Z})$, where $m$ is odd.

The abelianization of $\operatorname{SL}_2(\mathbb{Z}[\frac{1}{m}])$