O(G)=36 and o(H) =4 then H = Z(G)?

Question

Let $G$ be a group of order 36 and $H$ be a subgroup of $G$ with order 4. Then which of the following are true?

a) $H$ is a proper subgroup of $Z(G)$ (center of $G$)

b) $H = Z(G)$

c) $H$ is normal in $G$

d) $H$ is abelian

Solution:

$o(H) = 4$ therefore $H$ is abelian. $H$ is a $2$-sylow subgroup $n_2={1,3,9}$ therefore $H$ need not be normal always. Since $H$ is abelian It will be a subgroup of $Z(G)$. How to prove it is equal to $Z(G)$ or a proper subgroup of $Z(G)$?

Answer 1

Actually $H$ is not necessarily to be a subgroup of $Z(G)$.

$G = D_{36} = \langle r\rangle:\langle s\rangle \cong \mathbb{Z}_{18}:\mathbb{Z}_2$, a semi-direct product. Now $Z(G) = \{1,r^9\}$.