Let $t(G)$ denote the set of torsion elements of a group $G$. Prove that if $t(G)$ is a subgroup of G, then it is a normal subgroup of $G$.
Normality of the torsion part of a group given it is a subgroup
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group-theory
torsion-groups
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0By torsion element, do you mean element of finite order? Do you assume $G$ is abelian? I ask because in an arbitrary group $G$ the set of elements of finite order need not form a subgroup (e.g. $\langle a,b|a^2=b^2=1\rangle$ has elements of finite order $a,b$, but $ab$ has infinite order). – 2017-02-09
1 Answers
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Let $a\in t(G)$ then $\forall b \in G$ I show that $a^b \in t(G)$. Since $t(G)$ is a subgroup this makes it normal.
Let $n$ be the order of $a$, since conjugation is an automorphism: $1=(a^n)^b=(a^b)^n$. So, the order of $a^b$ divides $n$ therefore it is finite.