Central Series and the Nilpotent p-group

Definition $H/K$ is a central factor of a series $\ldots \unlhd K \unlhd H \ldots \unlhd G$ if $K \unlhd G$ and $H/K \le Z(G/K)$.
Lemma In a series such as the above, H/K is central iff $[H,G] \le K$.
proof: ($\Rightarrow$) Suppose $H/K$ is a central factor so $K \unlhd G$ and $H/K \le Z(G/K)$ then for all $g \in G$, $h \in H$, $KgKh = KhKg$ so $ghK=hgK$ and $h^{-1}g^{-1}hgK=K$ whence $[h,g] \in K$ thus $[H,G] \le K$. ($\Leftarrow$) very similar.. unproved

Definition A series with all factors central is a central series.

Definition A group with a central series is nilpotent.

Definition A series with all factors abelian is an abelian series.

Definition A group with an abelian series is solvable. (This may be familiar from Galois theory)

Definition A $p$-group is one whose order is a power of the prime $p$.

Lemma If $G$ is a nontrivial $p$-group then $Z(G)$ is also non-trivial.
proof: Each conjugacy class has size some power of $p$ and by the Conjugacy Class Formula the sum of these is $|G|$. The identity lies in a class of size $1$ so there must be at least $p-1$ other such conjugacy classes in $Z(G)$.

Theorem A group of order $p^2$ is abelian.
proof: A nontrivial element of $Z(G)$ either generates the whole group $C_{p^2}$ or generates $C_p \unlhd G$ in which case some element of $G$ not in that $C_p$ will generate the rest of group and $G$ is $C_p \times C_p$.

This same argument doesn't work for $p^3$ or higher: note $D_8$ is non-abelian, it's center is $C_2$ not the whole group.  If $G$ is abelian then $Z(G)=G$.

Theorem A $p$-group is nilpotent.
proof: Let $G$ be a $p$-group and we will proceed by induction on $|G|$. Let $Z = Z(G)$. So $G/Z(G)$ has a central series which by Noether3,4 is, up to isomorphism, of the form $$Z/Z = G_0/Z \unlhd G_1/Z \unlhd \cdots \unlhd G_n/Z = G/Z$$ with $(G_i/Z)/(G_{i-1}/Z)$ central (i.e. a subgroup of $Z((G/Z)/(G_{i-1}/Z))$) therefore (by Noether4) $G_i/G_{i-1} \le Z(G/G_i)$ and $$Z = G_0 \unlhd G_1 \unlhd \cdots \unlhd G_n = G$$ is a central series for our $p$-group if you stick 1 on the left.

• wikipedia.org/P-group#cite_note-1