# 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 $G$?

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?

## 1 Answer

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 $$h$$.