# Generalization of a property of proper conjugate subgroups of a finite nilpotent group

Let $G$ be a finite nilpotent group and $H, K$ be two proper non-maximal subgroups of $G$ such that $H\not\leq\Phi(G), K\not\leq\Phi(G)$ and $\langle H, K\rangle < G$.

Does there exist a maximal subgroup $M$ of $G$ such that $H\not\leq M$ and $K\not\leq M$? $\Phi(G)$ denotes the Frattini subgroup of $G$, which is the intersection of all maximal subgroups of $G$.

(If $H$ and $K$ are conjugate in $G$, then such $M$ exists. Because maximal subgroups of a nilpotent group are normal in it).

Sorry, the answer is no. In my previous "solution", I said that we could assume that $G$ is a $p$-group, but that is wrong.

Let $G = \langle a \rangle \times \langle b \rangle \times \langle c \rangle$, with $|a|=|b|=2$ and $|c|=3$ and $H = \langle a \rangle$, $K = \langle c \rangle$. So $G$ is a direct product of elementary abelian $p$-groups and $\Phi(G)$ is trivial.

All maximal subgroups of $G$ contain $\langle a,b \rangle$ or $\langle c \rangle$, so they all contain $H$ or $K$.

The answer is yes if $G$ is a $p$-group.

Since $\Phi(G) \le M$ for any maximal subgroup $M$ of $G$, we have $H \le M \Leftrightarrow \Phi(G)H \le M$, and so we can assume that $\Phi(G) < H$, and similarly $\phi(G) < K$. So, by working modulo $\Phi(G)$, we can assume that $\Phi(G)=1$.

If $G$ is a $p$-group, then it is elementary abelian. Let $1 \ne x_1 \in H$. If $x_1 \in K$, then we can take $M$ to be a maximal subgroup of $G$ not containing $x_1$, so assume that $x_1 \not\in K$. Similarly we can assume that there is $x_2 \in K \setminus H$. Then extend $\{x_1,x_2\}$ to a minimal generating set $\{x_1,x_2,\ldots,x_n\}$ of $G$ and take $M= \langle x_1x_2,x_3,\ldots,x_n\rangle$.

This does not use the assumption that $H$ and $K$ are non-maximal.

• Dear professor Holt, thank's a lot for the time on the question. Please let me know why do you assume that $H$ and $K$ are both maximal? – sebastian Sep 4 '14 at 9:36
• That ensures that $H \not\le M$ and $K \not\le M$. – Derek Holt Sep 4 '14 at 11:15
• Again I do apologize professor Holt! My mean is that why do you use the maximality of $H$ and $K$, while according to our assumption they are non-maximal? (May be $G$ is an elementary abelian group of rank greater than 2). – sebastian Sep 4 '14 at 12:53
• Dear professor Holt; thank you very much for your nice example! If it is possible, please explain a little about the validity of the property when "$G$ is a $p$-group". – sebastian Sep 4 '14 at 18:07
• I have added some explanation for the $p$-group case, which is hopefully correct this time! – Derek Holt Sep 4 '14 at 19:05