24 дек. 2017 г. · Lagrange's theorem is completely the same for infinite groups. Lagrange's theorem: let G be a group and H a subgroup of G. Then |G|=|H|[G:H].

26 февр. 2023 г. · Yes, it holds for all groups and subgroups. The finiteness assumption is not needed. Although it is not that useful for infinite groups. In ...

14 нояб. 2019 г. · Infinite groups are not ordered (without additional structure), so it doesn't make sense to talk about ordinals in this context; only cardinals, ...

17 февр. 2015 г. · By Lagrange's theorem, a group G for some H⩽G is partitioned into [G:H] many subsets, each with cardinality |H|. If |H| is finite, then the index of H must be

21 мар. 2021 г. · When I have a finite group G and some subgroup H, Lagrange's Theorem tells me that the number of cosets of H in G is |G||H|.

24 февр. 2023 г. · By definition [G:H] is the size of the class consisting of distinct costs of H in G . For infinite group it can be extended upto cardinality.

26 июн. 2020 г. · For any subgroup H of a group G we have |G|=|H|⋅|G/H|, where we are using products of possibly infinite cardinals.

10 янв. 2018 г. · Lagrange's theorem implies that all subgroups are of p-power order. It does not imply that all possible subgroup sizes are achieved, ...

9 сент. 2013 г. · A subgroup H of G is proper if H≠G and H is not the trivial group. So, if G is an infinite group, then G is a subgroup of G that is not proper; ...

18 дек. 2016 г. · Yes, this works perfectly fine. A ring is in particular a group with respect to the addition operation, and any ideal is in particular a subgroup.