FZ implies generalized subnormal join property

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., FZ-group) must also satisfy the second group property (i.e., group satisfying generalized subnormal join property)
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Statement

Suppose $G$ is a FZ-group (i.e., its center has finite index). Then, $G$ is also a group satisfying generalized subnormal join property: an arbitrary join of subnormal subgroups of $G$ is subnormal.

Facts used

1. Subnormality satisfies image condition
2. Finite implies generalized subnormal join property
3. Joins commute with images, i.e., the image of a join of subgroups is the join of their individual images.
4. Subnormality satisfies inverse image condition
5. Central factor implies normal

Proof

Given: A FZ-group $G$, a collection of subgroups $\{H_i{\}}_{i\in I}$, all subnormal in $G$. $H=⟨H_i{⟩}_{i\in I}$ is the join of the subgroups.

To prove: $H$ is subnormal in $G$.

Proof: Let $Z(G)$ denote the center of $G$.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	All the groups $H_{i}Z(G)/Z(G)$, viewed as subgroups of the quotient group $G/Z(G)$, are subnormal subgroups.	Fact (1)	$H_i$s are all subnormal in $G$		given-cum-fact direct.
2	$HZ(G)/Z(G)$ is subnormal in $G/Z(G)$.	Facts (2), (3)	$G$ is FZ, i.e., $G/Z(G)$ is finite.	Step (1)	Step-fact-given combination direct.
3	$HZ(G)$ is subnormal in $G$.	Fact (4)		Step (2)	Fact-step direct.
4	$H$ is normal in $HZ(G)$.	Fact (5)			If $K=HZ(G)$, then we have that $Z(G)\le C_K(H)$. Thus, $H{C}_K(H)=K$, so $H$ is a central factor of $K$, hence is normal in $K=HZ(G)$.
5	$H$ is subnormal in $G$.			Steps (3), (4)	Direct from steps and the definition of subnormal.