Some Properties of p-Groups

Theorem 1.1 If $\mid {\sl\large G} \mid = p^2$ for some prime p, then P is abelian.

proof:

$Z({\sl\large P}) \ne 1$ therefore ${\sl\large P}/Z({\sl\large P})$ is cyclic. Let x and $y \in {\sl\large G}$ and $\mid x \mid = \mid y \mid = p$ and choose $y \in {\sl\large P} - <x>$. Since $\mid <x,y> \mid$ > $\mid <x> \mid \Longrightarrow {\sl\large G} =$<x,y>. Clearly, $<x> \times <y> \cong Z({\sl\large P}) \times Z({\sl \large P})$ and so the mapping $\phi:(x^{a},y^{b}) \longrightarrow x^{a}y^{b}$ is an isomorphism. But $Z({\sl\large P}) \times Z({\sl \large P})$ is an abelian group and isomorphisms preserve structure. Thus, ${\sl\large G}$ must be an abelian group. Therefore ${\sl\large G}$ is not simple because it must have a subgroup of order p and all subgroups must be normal in an abelian group. No groups of order p² are simple.


Theorem 1.2 A group ${\sl\large G}$ with order equal to pq for pq primes and p < q contains a normal Sylow q-subgroup.

It follows that no group with order pq is simple. The proof follows directly from Sylow's Theorem. Sylow q-subgroups exist and $n_q \mid p$. Therefore, n_q = 1 and ${\sl\large G}$ contains a normal Sylow q-subgroup.


Theorem 1.3 If ${\sl\large H}$ is a proper subgroup of ${\sl\large G}$, then ${\sl\large H}$ is contained in ${\sl\large N}_G({\sl\large H})$.

proof:

If ${\sl\large Z}({\sl\large G}) \not\le {\sl\large H} \Longrightarrow {\sl\large Z}({\sl\large G}) \le {\sl\large N}_G({\sl\large H}).$

Therefore, ${\sl\large N}_G({\sl\large H}) \supset {\sl\large H}$.

Suppose that ${\sl\large Z}({\sl\large G}) \le {\sl\large H}$. We will prove this case by induction on the order of ${\sl\large G}$. If $\mid {\sl\large G} \mid = p^2$then ${\sl\large G}$ is abelian and ${\sl\large N}_G({\sl\large H}) = {\sl\large G}.$ Assume that for all groups with order $p^{\alpha - 1}$that ${\sl\large N}_G({\sl\large H}) \supseteq {\sl\large H}$.

Let $\mid {\sl\large G} \mid = p^{\alpha}$. ${\sl\large Z}({\sl\large G}) \ne 1$ and so form the factor group ${\sl\large G}/{\sl\large Z}({\sl\large G})$. The subgroup ${\sl\large H}$ maps to a subgroup of the factor group. But the order of the factor group ${\sl\large G}/{\sl\large Z}({\sl\large G})$ < $p^{\alpha}$. So the normalizer in the factor group contains the subgroup mapped by ${\sl\large H}$, and this subgroup correlates to ${\sl\large N}_G({\sl\large H})$ So by induction ${\sl\large N}_G({\sl\large H}) \supseteq {\sl\large H}$.

Theorem 1.4 (Frattini Argument) Let ${\sl\large H} \triangleleft {\sl\large G}$ and ${\sl\large P} \in Syl_{p}({\sl\large H})$. Then, ${\sl\large G} = {\sl\large H}{\sl\large N}_G({\sl\large P})$ and $\mid {\sl\large G}:{\sl\large H} \mid$ divides $\mid {\sl\large N}_G({\sl\large P}) \mid$.

proof:
${\sl\large H} \triangleleft {\sl\large G} \longrightarrow {\sl\large H}{\sl\large N}_G({\sl\large P}) \le {\sl\large G}$. Let $g\in {\sl\large G}$. Then $g{\sl\large P}g^{-1} \subseteq g{\sl\large H}g^{-1} \subseteq {\sl\large H}$. So both ${\sl\large P}$ and $g{\sl\large P}g^{-1} \subset {\sl\large H}$. By Sylow's Theorem, there exists an $x \in {\sl\large H}$ such that $x{\sl\large P}x^{-1} = g{\sl\large P}g^{-1}$. Therefore,

$$\begin{eqnarray*}x^{-1}g & \in & {\sl\large N}_G({\sl\large P})\\ g & \in & x{\... ... P})\\ g & \le & {\sl\large H}{\sl\large N}_G({\sl\large P})\\ \end{eqnarray*}$$

and ${\sl\large G} \le {\sl\large H}{\sl\large N}_G({\sl\large P})$.

Theorem 1.5 If ${\sl\large G} \in Syl_{p}(S_k)$ then ${\sl\large G} \in Syl_{p}(A_k)$. Furthermore, $\mid {\sl\large N}_{A_k}({\sl\large G}) \mid = \frac{1}{2}\mid {\sl\large N}_{S_k}({\sl\large G}) \mid$.

Lemma: If ${\sl\large G}$ is simple and ${\sl\large G} \le {\sl\large S}_{k}$ then ${\sl\large G} \le {\sl\large A}_{k}$.

proof:

Let ${\sl\large G} \in Syl_{p}({\sl\large A}_{k})$. By lemma ${\sl \large G} \le {\sl\large A}_{k} \Longrightarrow {\sl\large G} \in Syl_{p}({\sl\large A}_{k})$. Frattini's Argument gives that

$${\sl \large S}_{k} = {\sl\large N}_{s_k}({\sl\large G}){\sl\large A}_{k}$$

and so ${\sl\large N}_{S_k}({\sl\large G}) \not\subseteq {\sl\large A}_{k}$. Therefore ${\sl\large N}_{S_k}({\sl\large G}) \bigcap {\sl \large A}_{k}$ (which is the same as ${\sl\large N}_{A_k}({\sl\large G})$) must be a subgroup of index 2 in ${\sl\large N}_{S_k}({\sl\large G})$, because of the following

$$\begin{eqnarray*}\mid {\sl\large N}_{s_k}({\sl\large G}){\sl\large A}_{k} \mid &... ...arge N}_{s_k}({\sl\large G}) \bigcap {\sl\large A}_{k} \mid} \\ \end{eqnarray*}$$

The proof is complete.


Theorem 1.6 Let G be a group of order $p^{a}, a \ge 1$ Then if H is a normal subgroup of G then H contains a normal subgroup of every divisor p^b of $\mid H \mid$. In particular, G has a normal subgroup of every order p^b for $b \in \{ 1,2,3,...,a\}$.


In order to prove this we introduce the following lemma: $H \bigcap Z(G) \ne 1$. Therefore, by Cauchy's Theorem, $H \bigcap Z(G)$ contains a normal subgroup Z^* of Z(G) of order p.


proof:


If the order of G is p² we've already proved this. So assume that this holds true for all $2 \le b < a$. Choose G with order p^a. Z(G) commutes with everything so it is normal in G. Because $Z(G) \triangleleft G$ we can form the factor group G/Z(G) which has order $p^{b}, b \le a$. By induction, this factor group has normal subgroups, call them Z^*_i for all i that divide $\mid G/Z(G) \mid$. But the pre-image of each subgroup Z^*_(G)i is normal in G. Therefore, G has a normal subgroup for every order $p^{b}, 1 \le b \le a$.


The proof is complete.