Exercise 8.1.5

Proof.

(a)

Consider the right action of $G$ on itself defined by conjugation with $x^g=g^{-1}\mathit{xg}$, then $G$ in the disjoint union of the associate orbits:

$$G=\underset{x\in S}{\bigcup }{O}_x,$$

where $S$ is a complete set of representative for the conjugacy classes: for all $y\in G$, $|O_y\cap S|=1$.

The stabilizer of $x$ is the set of $g\in G$ such that $g^{-1}\mathit{xg}=x$, so $\mathit{xg}=\mathit{gx}$: this is the normalizer $C_x$ of $x$.

Consequently $|O_x|=(G:C_x)$, and so

$$|G|=\underset{x\in S}{\sum }(G:C_x).$$

Note that

$$(G:C_x)=1⟺C_x=G⟺\forall <!--\; FUNCTION\; APPLICATION\; -->g\in G,\mathit{gx}=\mathit{xg}⟺x\in Z=Z(G).$$

If we take apart these elements in the preceding sum, we obtain (noting that $Z\subset S$ since $O_x=\{x\}$ for all $x\in Z$)

$$|G|=\underset{x\in S}{\sum }(G:C_x)=\underset{x\in Z}{\sum }(G:C_x)+\underset{x\in S-Z}{\sum }(G:C_x)=|Z|+\underset{x\in S-Z}{\sum }(G:C_x).$$

Writing $T=S-Z$, we obtain so the class formula

$$|G|=|Z|+\underset{x\in T}{\sum }(G:C_x).$$

If $G$ is a $p$-groupe of order $p^n$, then for all $x\in S-Z$, $(G:C_x)>1$ is a power of $p$, so $(G:C_x)=p^k,k\ge 1$, thus $p$ divides $(G:C_x)$ for all $x\in T$.

As $p$ divides also $|G|$, the class formula implies that $p$ divides $|Z|\ge 1$, and so $|Z|\ge p$. Therefore the center of a $p$-group is not trivial.

(b)

If $g\in Z(G)$, then for all $x\in G$, and for all $k\in \mathbb{Z}$, $g^k\in Z$, so $x{g}^k=g^{k}x$, $x{g}^k{x}^{-1}=g^k\in Z$, therefore $⟨g⟩$ is normal in $G$.

(c)

If $n=1$, every group of order $p^n=p$ is cyclic, a fortiori solvable.

Using induction, suppose that all groups of order $p^k,k<n$ are solvable. Let $G$ be a group of order $p^n$.

Fix $g\in Z,g\ne e$. This is possible since $Z$ is not trivial. By part (b), $H=⟨g⟩$ is a normal cyclic subgroup $G$, so $H$ is solvable, and $G∕H$ as for cardinality a factor of $p^n$, so $|G∕H|=p^k$, with $k<n$ since $|H|>1$. The induction hypothesis implies that $G∕H=G∕⟨g⟩$ is solvable.

As $H$ and $G∕H$ are solvable, by Theorem 8.1.4, $G$ is also solvable, and the induction is done.

Every finite $p$-group is solvable.