# $\{ a \mid a\not \in [G,G],\ a \in Z(G) \} \cup \{e\}$ 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?

• Why are you moving to vector spaces? I thought we were discussing groups. Commented Jul 13, 2022 at 14:07
• I was thinking for an analogous case. Commented Jul 14, 2022 at 5:50

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.

• Pedantic note: except when $G$ is abelian. Commented Jul 13, 2022 at 21:58
• @MeesdeVries your note was not pedantic at all. In fact, non-abelian $[G,G]$ not containing $Z(G)$ was not enough either. I need exactly the consequence I claimed. Edited. Commented Jul 13, 2022 at 22:01
• Thanks for the answer. Commented Jul 14, 2022 at 5:49
• Should be "non-trivial proper subgroup"?
– spin
Commented Sep 11, 2022 at 13:16
• @spin If it's trivial then yes, the set in the OP's question would be $Z(G)$. Edited. Commented Sep 11, 2022 at 15:32

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.

• thanks for your answer. Its really nice. Commented Jul 14, 2022 at 5:49

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: $$G = C_2 \times D_8$$.