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

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  • $\begingroup$ 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
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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 $j \leq i$. $\square$

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

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