"Assume $P$ is a $p$-group and $N$ is normal in $P$ with this property that $P/N$ is an abelian elementary group. Prove that $\Phi(P)$ is in $N$. (Note: $\Phi(P)$ is intersection of maximal subgroups of $P$)"
I am going to bring my proof, which I am not sure about it.
Lemma 1: every finite abelian $p-$groups are in the form of $ G=Z_p \times \dots \times Z_p$.
Proof: We assume that $|G|=p^n$. By using induction, and the fact that each abelian group can be written as form of $ G=Z_{p^{k1}}\times \dots \times Z_{p^{ki}}$. If the group is cyclic then it should be $G=Z_p$.(each element should have the same order $p$ in elementary $p-$groups) So, we can assume it is not cyclic and each of $p^{ki}<p^{n}$, and this means that we have the induction assumption on them. Thus, we will reach the result. We need to check that the lemma is true for $|P|=p$ and $|P|=p^2$. We know that each $p^2$ groups are the form of $Z_{p^2}$ or $Z_{p} \times Z_{p} $. But, the cyclic form is not possible as it is an elementary group.
Lemma 2: if $G$ an abelian elementary group with the order of more than $p^2$, then its Frattini subgroup is $1$.
Proof: Each elementary group is in the form of $ G=Z_p \times \dots \times Z_p$(by using Lemma1). I was thinking that if we drop one of $Z_p$ then the remaining will be a maximal subgroup in the $G$(Am I right here???). Thus, their intersection will be 1, and $G$ has a trivial Frattini subgroup.
Now we consider $G=P/N$, and it is an abelian elementary group. if $|P/N|=p$ then $N$ itself is a maximal subgroup in $P$, and from the definition the $\Phi(P)$ is in the $N$. So, we can assume that $|P/N| \ge p$. Now we want to assume that $\Phi(P)$ is not fully in $N$, so there is $x \in \Phi(P)$ and $x \notin N$. So, $xN \neq 1N$ in $P/N$. Moreover, we know that in $P/N$ all its nongenerator is 1( as its Frattini subgroup is only 1). So $xN$ in $P/N$ is a generator. There should exist $I/N$ (not whole $P/N$) a subgroup in $P/N$ that: $$ \langle xN,I/N \rangle =P/N$$ In the $aN$, $a$ an arbitrary element in $P$: $$ \langle I,x \rangle \cap \; aN \neq \emptyset \quad \text{because we have}\; aN \; \text{in} \langle xN,I/N \rangle \text{as an element} $$ But, we have $N \subseteq I$. We have $an \in \; \langle x,I \rangle $ and $n^{-1} \in I$, Thus: $$an.n^{-1}=a \in \; \langle x,I \rangle$$. But, $I \neq P$ ,$x$ a nongenrator, and $ \langle x , I \rangle =P$. Contradiction.
My further requests:
- I am new in using Latex every points will be a great help.
- Is there any rules in writing mathematics' proofs? I think my writing is not organized.