I know that t(G) is the set of all elements of G of finite order. I've managed to show that if G is abelian then t(G) is a subgroup, but I am stuck trying to show t(G) is a normal subgroup if t(G) is a subgroup, without using the abelian property. Any help would be appreciated!!
Torsion part of a group G
-3
$\begingroup$
linear-algebra
finite-groups
normal-subgroups
2 Answers
1
If $g$ has finite order then so does every conjugate of $g$. Thus the torsion sub set is preserved by conjugation. Of course, there is no reason to suppose that the torsion subset is a subgroup.
1
I'm confused why you want to avoid using the fact that $G$ is abelian; without the assumption that $G$ is abelian, $t(G)$ might not even be a subgroup at all! For example, consider the group $\langle a, b: a^2=b^2=e\rangle$. The elements $a$ and $b$ are each torsion, but $ab$ is not.
That said, it can be done. Suppose $g\in t(G)$ and $a\in G$; then $(aga^{-1})^n=ag^na^{-1}$, so if $g$ has order $k$ then $(aga^{-1})^k=e$ and so $aga^{-1}\in t(G)$. Note that this shows that $t(G)$ is always closed under conjugation, even if $t(G)$ is not a group!