Class two implies generated by abelian normal subgroups

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., group of nilpotency class two) must also satisfy the second group property (i.e., group generated by abelian normal subgroups)
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Statement

Suppose $P$ is a group of nilpotency class two. Then, $P$ is a group generated by abelian normal subgroups. In fact, $P$ is a union of abelian normal subgroups.

Facts used

Proof

Given: A group $P$ with center $C$ such that $G:=P/C$ is an abelian group.

To prove: $P$ is a union of abelian normal subgroups of itself.

Proof:

Step no.	Assertion	Given data used	Facts used	Proof steps used	Explanation
1	$G$ is a union of cyclic subgroups $G_{i},i\in I$	--	fact (1)	--
2	Each $G_{i}$ is normal in $G$	$G$ is abelian	fact (2)	step (1)
3	Let $P_{i}$ be the inverse image (for the quotient map $P\to G$ with kernel $C$ ) in $P$ of $G_{i}$ . Then, each $P_{i}$ is abelian.	$C$ is the center of $P$	fact (3)	step (1)	[SHOW MORE] We have $P_{i}/C\cong G_{i}$ , which by step (1) is cyclic. Thus, $P_{i}$ is a cyclic extension of the central subgroup $C$ of $P$ , hence is abelian.
4	Each $P_{i}$ is normal in $P$	--	fact (4)	step (3) (construction of $P_{i}$ s)
5	$P$ is the union of the $P_{i}$ s	--	--	steps (1) and (3)	[SHOW MORE] By step (1), every element of $G$ lies in one of the $G_{i}$ s. Since every element of $P$ maps to some element of $G$ , hence its image mod $C$ lies in one of the $G_{i}$ s. Thus, it lies in the corresponding $P_{i}$ .