# Higher centers are characteristic.

Let $G$ be a group. Define $\zeta^i=\zeta^i(G)$ inductively as follows: $\zeta^0=1$and $\zeta^{i+1}$ is the subgroup of $G$ for which $$\frac{\zeta^{i+1}(G) }{\zeta^i(G)}=Z\left(\frac{G}{\zeta^i(G)}\right)$$

Thus $\zeta^{i+1}(G)=\{g\in G:\forall g'\in G,[g,g']\in \zeta^i(G)\}$ is the largest$^{1}$ subgroup of $G$ for which $[\zeta^{i+1}(G),G]\leqslant \zeta^i(G)$. My questions are two:

$(1)$ Is is true the last inclusion is an equality for each $i=0,1,2,\ldots$? It is easily seen the last definition agrees for $i=0$, for $Z(G)$ is the largest subgroup of $G$ for which $[Z(G),G]\leqslant 1$, i.e. $[Z(G),G]=1$, but I am not sure if we have equality for $i=1,2,\ldots$.

$(2)$ I want to show each $\zeta^{i}(G)$ is characteristic. $\zeta^{0}(G)=1$ is trivially characteristic. Now assume $\zeta^{i}(G)$ is, and observe that if $\eta$ is an automorphism of $G$, then $[\zeta^{i+1}(G),G]\leqslant \zeta^{i}(G)$ becomes $[\eta\zeta^{i+1}(G),G]\leqslant \zeta^{i}(G)$ for both $G$ and $\zeta^{i}(G)$ are characteristic, and $\eta[H,K]=[\eta H,\eta K]$. This means$^{1}$ $$\frac{\eta\zeta^{i+1}(G)}{\zeta^i(G)}\leqslant \frac{\zeta^{i+1}(G) }{\zeta^i(G)}=Z\left(\frac{G}{\zeta^i(G)}\right)$$

so $\eta \zeta^{i+1}(G)\leqslant \zeta^{i+1}(G)$ and $\zeta^{i+1}(G)$ is characteristic. Is there a better proof?

$1.$ Lemma If $K\lhd G$, $K\leqslant H\leqslant G$ , then $[H,G]\leqslant K\iff H/K\leqslant Z(G/K)$. Thus if $H$ is some group above $\zeta^i(G)$ for which $[H,G]\leqslant \zeta^i(G)$ the lemma gives $H/ \zeta^i(G)\leqslant \zeta^{i+1}(G)/\zeta^i(G)$, which implies $H\leqslant \zeta^{i+1}(G)$.

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For (2): If $H$ is characteristic in $G$ and $K$ is characteristic in $G/H$ then the preimage of $K$ in $G$ (by the quotient map) is characteristic in $G$. – Tobias Kildetoft Mar 7 '14 at 8:11

(1): Consider $G = D_8 \times C_2$, where $D_8$ is dihedral of order $8$ and $C_2$ is cyclic of order $2$. For this group, $\zeta^2 = G$ and $[G,G]$ is a proper subgroup of $\zeta^1$.

(2): If $N$ is characteristic in $G$, then we have the map $\operatorname{Aut}(G) \rightarrow \operatorname{Aut}(G/N)$ defined by $\phi \mapsto \hat{\phi}$, where $\hat{\phi}(xN) = \phi(x)N$. Therefore if $H/N$ is characteristic in $G/N$, then $H$ is characteristic in $G$.

With this, you can prove (2) using the fact that the center of any group is characteristic.

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Is there any particular reason you chose $D_8 \times D_4$, rather than just $D_8$ or $D_4$? As the reasoning in my answer shows, any nilpotent group would give an example – zcn Mar 7 '14 at 8:58
@user115654: For some reason I was only thinking about nilpotent groups and the upper central series being finite: $\zeta^0, \zeta^1, \ldots, \zeta^c = G$.. so your easy example didn't cross my mind. Note that when $G = D_{2^n}$, we do have equality $[\zeta^{i+1}, G] = \zeta^i$ for $0 \leq i < n-1$, and $\zeta^{n-1} = G$. – Mikko Korhonen Mar 7 '14 at 9:11

The inclusion in $(1)$ cannot be an equality for every $i$, in general. If $G$ is nilpotent, then $G = \zeta^i(G)$ for some $i$, and also $[G, G] \subsetneq G$, so $[\zeta^{i+1}(G), G] \subseteq [G, G] \subsetneq \zeta^i(G)$.

A nice way to see $(2)$ has already been addressed in the comments above.

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