Center of $G/Z(G)$.

Question

Question. Prove or disprove: The center of $G/Z(G)$ is trivial for all finite groups $G$. (Here $Z(G)$ denotes the center of $G$.)

Attempt at a Proof. Assume the contrary that $G/Z(G)$ has a non-trivial center.

Let $g\notin Z(G)$ be such that $gZ(G)$ is in the center of $G/Z(G)$.

Then $$\boxed{xgx^{-1}\in gZ(G)}$$ for all $x\in G$.

We note that $xgx^{-1}=ygy^{-1}$ if and only if $xC_G(g)=yC_G(g)$, where $C_G(g)$ denotes the centralizer of $g$ in $G$.

Since $g\notin Z(G)$, we have $G\neq C_G(g)$.

I couldn't go any further.

Can someone help?

Answer 1

Do you know this result? $$G/Z(G) \cong \operatorname{Inn}(G)$$

You may want to look at $S_n$.

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Edit: It fails for $n=6$, but it's not easy to see why $\{e\} <\operatorname{Inn}(S_6) < \operatorname{Aut}(S_6)$. But $S_6$ is a good counter-example.

To study more about $S_6$ and why $\{e\} < \operatorname{Inn}(S_6) < \operatorname{Aut}(S_6)$ read this wikipedia entry.

Answer 2

Consider $D_4$ (dihedral group or the symmetry group of a square). It has 8 elements and can be represented as a permutation group on vertices $\{1,2,3,4\}$ of a square as follows:

• $(1,2,3,4),(1,3)(2,4),(1,4,3,2),(1)(2)(3)(4)$ - rotations;
• $(2,4)(1)(3),(1,3)(2)(4),(1,2)(3,4),(1,4)(2,3)$ - symmetries.

So, $Z(D_4)=\{(1,3)(2,4),(1)(2)(3)(4)\}$ of order 2. Thus $|D_4/Z(D_4)|=4$ and the factor-group is Abelian (as all groups of order less then 6).

Answer 3

Recall that if $G$ is a non-abelian group of order $p^3$, then $\overline G=G/Z(G)\simeq C_p\times C_p$. Then $Z(\overline G)=\overline G$ will be nontrivial.

Answer 4

Not only the statement that ”$G/Z(G)$ is trivial” is false, but a very important class of groups is defined by means of “repeated centers”. Let $G$ be a group and define $$ Z_0(G)=\{1\},\qquad Z_{k+1}(G)/Z_k(G)=Z(G/Z_{k}(G))\text{ for $k\ge0$}. $$ In other words, $Z_{k+1}(G)$ is the inverse image of $Z(G/Z_{k}(G))$ under the canonical projection $G\to G/Z_{k}(G)$.

A group is called nilpotent if there is a $k$ such that $Z_k(G)=G$.

Nilpotent groups are abundant: for instance, every finite $p$-group is nilpotent because of the following lemma, whose easiest proof relies on the study of conjugacy classes.

Lemma. Every finite $p$-group $G\ne\{1\}$ has nontrivial center.