The lower exponent $p$-central series for a $p$-group $G$ is defined by $G=P_1(G) > P_2(G) > \ldots > P_c(G) = 1$, where $$P_i(G)=[P_{i-1}(G), G] P_{i-1}(G)^p.$$ If $G_i=G/P_i(G)$ and $A:G_{i+1} \to G_i : g P_{i+1}(G) \mapsto g P_i(G)$, then $\ker(A) = P_i(G)/P_{i+1}(G)$.

Notice $\ker(A) \leq G_{i+1}$. Is $\ker(A) = P_i(G)/P_{i+1}(G) \cong P_i(G_{i+1})$?

  • $\begingroup$ It is a vector space over Z/pZ, call it Vi. Its rank depends on G. There is a linear transformation from Vi to V(i+1) that takes a coset gP(i+1)(G) to the coset g^p P(i+2)(G), and a biinear transformation from Vi x Vj to V(i+j) taking cosets g x h to the coset of [g,h]. The direct sum of the Vi then becomes a “restricted Lie algebra.” The individual Vi are not too interesting by themselves. V1 is G/Phi(G), the Frattini quotient. $\endgroup$ – Jack Schmidt Mar 10 '14 at 15:09
  • $\begingroup$ Thnks Jack Schmidth for your brief comments, is it true that KerA=Pi(G)/Pi+1(G) is isomorhic to Pi(Gi+1) $\endgroup$ – Fazli Amin Mar 10 '14 at 15:24
  • $\begingroup$ Yes. Commutators and p'th powers work very predictably in quotient groups. $\endgroup$ – Jack Schmidt Mar 10 '14 at 15:32
  • $\begingroup$ Thanks, i need the proof of this isomorphism, i tried by myself but it did not works, if possible please send me its proof $\endgroup$ – Fazli Amin Mar 10 '14 at 15:43
  • $\begingroup$ if possible please snd me the proof that KerA=Pi(G)/Pi+1(G) is isomorhic to Pi(Gi+1) $\endgroup$ – Fazli Amin Mar 10 '14 at 16:13

This is a routine verification. Fix a prime $p$.

Proposition: $P_j(G)/N = P_j(G/N)$ for all groups $G$, positive integers $j$, and normal subgroup $N \unlhd G$ with $N \leq P_j(G)$.

Proof: This is clearly true for $j=1$ since both sides are $G/N = G/N$. Suppose by induction that $1 < j$ and $P_{j-1}(G)/N = P_{j-1}(G_i)$. Then $$P_{j}(G/N) = [ P_{j-1}(G/N), G/N ] P_{j-1}(G/N)^p = [ P_{j-1}(G)/N, G/N ] \left( P_{j-1}(G)/N\right)^p$$ Notice that $[H/N,K/N] = [H,K]N/N$ and $(H/N)^p = H^pN/N$ since commutators and quotients work predictably in quotients: $[hN,kN] =[h,k]N$ and $(hN)^p = h^p N$ by definition of multiplication in a quotient group. Hence $$P_{j}(G/N) = [ P_{j-1}(G), G ] \left( P_{j-1}(G)\right)^p N/N = P_j(G) N/N = P_j(G)/N. \qquad \square$$

Corollary: $P_j(G_i) = P_j(G)/P_i(G)$ for all groups $G$ and positive integers $1 \leq j \leq i$.

Proof: Take $N=P_i(G)$. Then $P_i(G) \leq P_j(G)$ since $j \leq i$. $\square$

  • $\begingroup$ Thanks allot, Jack Schmidt $\endgroup$ – Fazli Amin Mar 14 '14 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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