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If $G$ is a finite group and $H$ is a normal subgroup of $G$ of order $2$, then what can be the order of center of $G$ ?

a. $0$

b. $1 $

c. an even integer $≥ 2$

d. an odd integer $≥3$

Since $H$ is a normal subgroup of $G$, normalizer of $H$, $N(H) = G$ and center of $G$, $Z(G)$ is a subgroup of $N(H)$.

These facts do not lead to any result. How to solve this?

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Let $H=\{e,h\}$. Since $H $ is normal and only has two elements, we have $$ghg^{-1}=h\iff gh=hg,\; \forall g\in G.$$ Hence $h \in Z(G).$ Also note that the order of $h $ is $2$. Hence $Z(G) $ contains even number of elements(by Lagrange's) including $e $ and $h $.

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