Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $H\leqslant \zeta^{i+1}(G)$.

share|improve this question
3  
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 at 8:11

2 Answers 2

up vote 2 down vote accepted

(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.

share|improve this answer
    
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 at 8:58
1  
@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 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.