$\{ a \mid a\not \in [G,G],\ a \in Z(G) \} \cup \{e\}$ is a group or not Let $G$ be a group. I want to check whether the set
$$\{ a \mid a\not \in [G,G],\ a \in Z(G) \} \cup \{e\}$$
is a subgroup or not, where $[G,G]$ and $Z(G)$ are the commutator subgroup and center of the group $G$ respectively.

In a vector space $V$ we can find the complement subspace $U$ of any its subspace $U$ such that $V=W\oplus U$? Can we generalize this to my case?

 A: This is not true.
Let $G$ be any finite group where $[G,G]\cap Z(G)$ is a proper, non-trivial subgroup of $Z(G)$. (E.g., $C_2\times D_8$.) Thus it contains at most half of the members of $Z(G)$. Your subset, therefore contains more than half of the members of $Z(G)$ (as it also includes the identity), so cannot be a proper subgroup by Lagrange's theorem.
A: This is not true.
Take $G=\text{GL}_2(\mathbb{R})$ and denote your set by $A(G)$. Then we have $-2\cdot I_2,\frac{1}{2}\cdot I_2\in A(G)$ but their product is not in $A(G)$. Indeed, $[G,G]=\text{SL}_2(\mathbb{R})$ and the center of $G$ are the scalars, i.e. $Z(G)=\{\lambda\cdot I_2:\lambda\in \mathbb{R}\setminus\{0\}\}$.
Let me know if something is unclear.
A: The following is just an expanded version of David Craven's answer.
Denote $Z = Z(G)$, and $H = [G,G] \cap Z$. The set you are looking at is $(Z \setminus H) \cup \{e\}$.
In general, you would not expect this to be a subgroup of $Z$, for any subgroup $H$ of $Z$.
Here is one reason:

Let $G$ be any group, and let $M$ be a proper subgroup. Then $G \setminus M$ generates $G$.

So if $(Z \setminus H) \cup \{e\}$ were a subgroup, it would have to be equal to $\{e\}$ or $Z$. In other words, either $H = Z$ or $H = \{e\}$.
Thus for a counterexample to the claim, it suffices to find $G$ such that $\{e\} \lneq [G,G] \cap Z(G) \lneq Z(G)$.
One example, given in the other answer: $G = C_2 \times D_8$.
