📂Abstract Algebra

p-Groups in Abstract Algebra

Definition¹

Let $e$ be the identity of a finite group $G$. For the smallest $n \in \mathbb{N}$ such that $g \in G$ satisfies $g^{n} = e$, we denote $|g| = n$. When for every $g \in G$ and a given prime $p$ there exists an integer $m \ge 0$ satisfying $|g| = p^{m}$, $G$ is called a $p$-group.

Explanation

If $|G| = p^{m}$, then $G$ is a $p$-group, and the following theorem is known.

Theorem

Let $X_{G} : = \left\{ x \in X \ | \ gx = x , g \in G \right\}$ be the set unaffected by the action, which can be written as $\displaystyle X_{G} = \bigcap_{ g \in G} X_{g}$. If a finite group $G$ satisfies $|G| = p^{m}$ and $X$ is a $G$-set, then $$ |X| \equiv |X_{G}| \pmod{p} $$

Proof

Suppose $X$ has $r$ orbits. If we denote the elements picked from each orbit by $x_{1} , \cdots , x_{r}$, then $$ |X| = \sum_{i=1}^{r} | G x_{i} | $$ Letting $s = |X_{G}|$, we have $0 \le s \le r$ and $$ |X| = |X_{G}| + \sum_{i=s+1}^{r} | G x_{i} | $$

Property of isotropy subgroups: If $X$ is a $G$-set, then $|Gx| = ( G : G_{x})$. If $G$ is a finite group, then $|Gx|$ is a divisor of $|G|$.

Since $|G| = p^{m}$, its divisor $|Gx_{i}|$ must appear as a power of $p$. Therefore, for some $k \in \mathbb{Z}$, $$ |X| = |X_{G}| + p k $$ and by the definition of congruence, the following holds. $$ |X| \equiv |X_{G}| \pmod{p} $$

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