Showing an element is in the center of $G$, $|G| = 8$
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Let $G$ be a group of order $8$ and $x∈G$ such that order of $x$ is $4$. I want to show that $x^2∈Z(G)$.
Can anyone help?
abstract-algebragroup-theory
asked 2012-12-12
user id:52730
6
11bronze badges
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$G$ is a p-group and therefore is nilpotent. Hence, the center of $G$ has order at least $2$ and all normal subgroups of $G$ have at least two elements of $Z(G)$. All that remains to show is that $\langle x\rangle$ is a normal subgroup of $G$. However, this is trivial, since the index of that subgroup is the smallest prime dividing $|G|$. – 2012-12-12