Let G be any group and let H be the subgroup of G generated by all elements of finite order. Is there a name for H?
11
-
8$\begingroup$ @NoahSchweber I think torsion subgroup is only used when the set of torsion elements form a subgroup, which is not always the case. I am not aware of the subgroup generated by the torsion elements having any particularly interesting properties in general. $\endgroup$– Derek HoltCommented Nov 8, 2015 at 20:39
-
6$\begingroup$ @NoahSchweber The non-abelian case is quite different, since it is quite common for torsion elements to generate the whole group! $\endgroup$– Igor RivinCommented Nov 8, 2015 at 20:39
-
1$\begingroup$ I do not think torsion subgroup would usually be used in the case of non-Abelian groups since the set of elements of finite order in an infinite non-Abelian group is not usually a subgroup, so to call $H$ the torsion subgroup could be misleading. Note however, that the subgroup $H$ is a normal subgroup . Unlike the Abelian case, $G/H$ need not be torsion free in general. $\endgroup$– Geoff RobinsonCommented Nov 8, 2015 at 20:39
-
2$\begingroup$ More generally the torsion elements form a subgroup in nilpotent groups. $\endgroup$– Derek HoltCommented Nov 8, 2015 at 20:42
-
2$\begingroup$ In fact, $X(G)$ (to follow the notation of @YCor) is not just characteristic, but fully invariant, since given any endomorphism $\varphi\colon G\to G$, the image of an element of finite order must be of finite order, so $\varphi(X(G))\subseteq X(G)$. It is even more, since the construction is functorial, in that if $G$ and $H$ are groups, and $\varphi\colon G\to H$ is a group homomorphism, then $\varphi(X(G))\subseteq X(H)$, since the image of the generating set of $X(G)$ lies inside $X(H)$. $\endgroup$– Arturo MagidinCommented Nov 8, 2015 at 23:14
|
Show 6 more comments