Definition $\Gamma_1(G) = G$ and $\Gamma_{i+1}(G) = [\Gamma_i(G),G]$.
Definition $Z_0(G) = 1$ and $Z_i(G)$ is defined to be the unique (prove this) group sch that $Z_i(G)/Z_{i-1}(G) = Z(G/Z_{i-1}(G))$.
Note: This is the biggest possible group that $Z_i(G)$ could be and it still satisfying the condition needed for a central series.
Definition lower central series $$G = \Gamma_1(G) \unrhd \Gamma_2(G) \unrhd \Gamma_3(G) \unrhd \cdots.$$
Definition upper central series $$1 = Z_0(G) \unlhd Z_1(G) \unlhd Z_2(G) \unlhd \cdots.$$
Proposition $\Gamma_2(G) = G' = G^{(1)}$ and $Z_1(G) = Z(G)$ and all $Z$-terms are actually characteristic subgroups.
Theorem $G$ is nilpotent iff $\Gamma_n(G) = 1$ (for some n) iff $Z_n(G) = G$ (for some n), furthermore if $G$ had a central series $$1 = G_0 \unlhd G_1 \unlhd \cdots \unlhd G_r = G$$ then we have (the optimality conditions) for each $0 \le i \le r$, $\Gamma_{r-i+1}(G) \le G_{i} \le Z_i(G)$ and finally $\Gamma_{c+1}(G) = 1$ iff $Z_c(G) = G$.
proof: Certainly if either $\Gamma_{n+1}(G) = 1$ iff $Z_n(G) = G$ for some $n$ then the lower or upper series are a central series and $G$ is nilpotent. Now suppose $G$ is nilpotent and take a central series we'll prove the optimality conditions separately by induction.
First $\Gamma_{i+1}(G) \le G_{r-i}$, assume it's true for $i$. then $[\Gamma_{i+1}(G),G] \le [G_{r-i},G] \le G_{r-i-1}$ done.
Second $G_{i} \le Z_i(G)$, to get $G_{i+1} \le Z_{i+1}(G) = Z(G/Z_i(G))$ we'll show that $G_{i+1}/Z_i(G) \le Z(G/Z_i(G))$ by showing for arbitrary elements $l \in G_{i+1}$ and $g \in G$ that $lZ_i(g)$ commutes with $gZ_i(G)$... but $[G_{i+1},G] \le G_i \le Z_i(G)$ so we are done! Explaining this a bit more: for any $x \in G_{i+1}$ and $y \in G$ we have $[x,y] \in Z_i(G)$, that's the same as $x Z_i(G)$ commuting with $y Z_i(G)$, which is the same as $xZ_i(G)$ being in the center $Z(G/Z_i(G))$ i.e. $G_{i+1}/Z_i(G) \le Z(G/Z_i(G))$.
Finally suppose $\Gamma_{c+1}(G)=1$, by taking $G_j = \Gamma_{c-j+1}(G)$ for each $0 \le j \le c$ and $r=c$ we have $G = \Gamma_1(G) = G_c \le Z_c(G)$ so $g = Z_c(G)$. On the other hand if $Z_c(G) = G$ then by taking $G_j = Z_j(G)$ we have $\Gamma_{c+1}(G) \le G_0 = 1$ so $\Gamma_{c+1}(G) = 1$.
Definition The smallest $n$ such that $Z_n(G) = G$ is the nilpotency class.
todo: prove it's the same for gamma and and other series.
No comments:
Post a Comment