Return to Answer

If $G$ is a non-trivial finite group which can be generated by two non-trivial elements of prime power orders, then the answer to the question is affirmative. Let $\mathcal{G}$ be the set of all subgroups of $G$ and let $G=\langle x,y\rangle$, where $x$ and $y$ are non-trivial elements of $G$ of prime power oders. Assume $A=\lbrace H\in \mathcal{G} \;|\; x\in H, y\not\in H \rbrace$, $B=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\in H \rbrace$, $C=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\not\in H \rbrace$ and $E=\lbrace H\in \mathcal{G} \;|\; x\in H, y\in H \rbrace$. So $E=\{G\}$ and $|C|\geq 1$ as $\{1\}\in C$. It follows that either $|A|\leq |\mathcal{G}|/2$ or $|B|\leq |\mathcal{G}|/2$.

I think (if I am not wrong) that every (if not "most" of!) finite simple groups can be generated by two elements of prime power orders. So the question has affirmative answer for finite simple groups and $2$-generated groups of prime power orders.

UPDATE (16/January/2014): Inspired by the Russ Answer, one can show that the question has affirmative answer for all finite groups $G$. Let $\Phi(G)$ be the Frattini subgroup of $G$ which by definition is the intersection of all maximal subgroups of $G$. Note that $G\not=\Phi(G)$, as $G$ is finite and it has at least one maximal subgroup. If $g$ is any prime power order element of $G\setminus \Phi(G)$, then $G=M\langle g\rangle$ for some maximal subgroup $M$ of $G$ so that $g\not\in M$. Now the map $H\mapsto H\cap M$ is injective for all subgroups $H$ containing $g$. So we may assume that all prime power order elements are in $\Phi(G)$ which implies that $G=\Phi(G)$, a conradiction, since a finite group can be generated by its Sylow subgroups.

SECOND UPDATE(16/January/2014) I found a gap in the above proof (Update), why $G=M\langle g\rangle$? It may not hold in general. So the answer of the question is still open for myself in general. Sorry!!

THIRD UPDATE (16\January/2014): If one can prove that $H=\langle g,H\cap M\rangle$ for all subgroups $H$ containing $g$, we are done. Anyway the gap in the first UPDATE is serious, and all my UPDATES do not say more than what has been said by Russ.

If $G$ is a non-trivial finite group which can be generated by two non-trivial elements of prime power orders, then the answer to the question is affirmative. Let $\mathcal{G}$ be the set of all subgroups of $G$ and let $G=\langle x,y\rangle$, where $x$ and $y$ are non-trivial elements of $G$ of prime power oders. Assume $A=\lbrace H\in \mathcal{G} \;|\; x\in H, y\not\in H \rbrace$, $B=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\in H \rbrace$, $C=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\not\in H \rbrace$ and $E=\lbrace H\in \mathcal{G} \;|\; x\in H, y\in H \rbrace$. So $E=\{G\}$ and $|C|\geq 1$ as $\{1\}\in C$. It follows that either $|A|\leq |\mathcal{G}|/2$ or $|B|\leq |\mathcal{G}|/2$.

I think (if I am not wrong) that every (if not "most" of!) finite simple groups can be generated by two elements of prime power orders. So the question has affirmative answer for finite simple groups and $2$-generated groups of prime power orders.

UPDATE (17/January/2014): Inspired by Russ, one can show that the question has affirmative answer for finite groups having a maximal subgroup(not necessarily normal) of prime index. Let $M$ be a maximal subgroup of $G$ of prime index $p$. If there exists an element $g\in G\setminus M$ of $p$-power order, then $G=M\langle g\rangle$. The latter holds since $$|M\langle g\rangle|=\frac{|M||g|}{|M\cap \langle g\rangle|}.$$ Now the map $H\mapsto H\cap M$ is injective for all subgroups $H$ containing $g$. So we may assume that all $p$-power order elements are in $M$ which implies that $M$ contains all Sylow $p$-subgroups of $G$, a contradiction. This slightly improves the result of Russ, as we do not assume $M$ is normal in the group.

If $G$ is a non-trivial finite group which can be generated by two non-trivial elements of prime power orders, then the answer to the question is affirmative. Let $\mathcal{G}$ be the set of all subgroups of $G$ and let $G=\langle x,y\rangle$, where $x$ and $y$ are non-trivial elements of $G$ of prime power oders. Assume $A=\lbrace H\in \mathcal{G} \;|\; x\in H, y\not\in H \rbrace$, $B=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\in H \rbrace$, $C=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\not\in H \rbrace$ and $E=\lbrace H\in \mathcal{G} \;|\; x\in H, y\in H \rbrace$. So $E=\{G\}$ and $|C|\geq 1$ as $\{1\}\in C$. It follows that either $|A|\leq |\mathcal{G}|/2$ or $|B|\leq |\mathcal{G}|/2$.

I think (if I am not wrong) that every (if not "most" of!) finite simple groups can be generated by two elements of prime power orders. So the question has affirmative answer for finite simple groups and $2$-generated groups of prime power orders.

UPDATE (16/January/2014): Inspired by the Russ Answer, one can show that the question has affirmative answer for all finite groups $G$. Let $\Phi(G)$ be the Frattini subgroup of $G$ which by definition is the intersection of all maximal subgroups of $G$. Note that $G\not=\Phi(G)$, as $G$ is finite and it has at least one maximal subgroup. If $g$ is any prime power order element of $G\setminus \Phi(G)$, then $G=M\langle g\rangle$ for some maximal subgroup $M$ of $G$ so that $g\not\in M$. Now the map $H\mapsto H\cap M$ is injective for all subgroups $H$ containing $g$. So we may assume that all prime power order elements are in $\Phi(G)$ which implies that $G=\Phi(G)$, a conradiction, since a finite group can be generated by its Sylow subgroups.

SECOND UPDATE(16/January/2014) I found a gap in the above proof (Update), why $G=M\langle g\rangle$? It may not hold in general. So the answer of the question is still open for myself in general. Sorry!!

THIRD UPDATE (16\January/2014): We may fill the gap as follows:Since $G=\langle g, M\rangle$, it follows that $H=\langle g,H\cap M\rangle$, by Dedekind Modular Law. Now the reset of the proof is valid, I hope.

If $G$ is a non-trivial finite group which can be generated by two non-trivial elements of prime power orders, then the answer to the question is affirmative. Let $\mathcal{G}$ be the set of all subgroups of $G$ and let $G=\langle x,y\rangle$, where $x$ and $y$ are non-trivial elements of $G$ of prime power oders. Assume $A=\lbrace H\in \mathcal{G} \;|\; x\in H, y\not\in H \rbrace$, $B=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\in H \rbrace$, $C=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\not\in H \rbrace$ and $E=\lbrace H\in \mathcal{G} \;|\; x\in H, y\in H \rbrace$. So $E=\{G\}$ and $|C|\geq 1$ as $\{1\}\in C$. It follows that either $|A|\leq |\mathcal{G}|/2$ or $|B|\leq |\mathcal{G}|/2$.

I think (if I am not wrong) that every (if not "most" of!) finite simple groups can be generated by two elements of prime power orders. So the question has affirmative answer for finite simple groups and $2$-generated groups of prime power orders.

UPDATE (16/January/2014): Inspired by the Russ Answer, one can show that the question has affirmative answer for all finite groups $G$. Let $\Phi(G)$ be the Frattini subgroup of $G$ which by definition is the intersection of all maximal subgroups of $G$. Note that $G\not=\Phi(G)$, as $G$ is finite and it has at least one maximal subgroup. If $g$ is any prime power order element of $G\setminus \Phi(G)$, then $G=M\langle g\rangle$ for some maximal subgroup $M$ of $G$ so that $g\not\in M$. Now the map $H\mapsto H\cap M$ is injective for all subgroups $H$ containing $g$. So we may assume that all prime power order elements are in $\Phi(G)$ which implies that $G=\Phi(G)$, a conradiction, since a finite group can be generated by its Sylow subgroups.

SECOND UPDATE(16/January/2014) I found a gap in the above proof (Update), why $G=M\langle g\rangle$? It may not hold in general. So the answer of the question is still open for myself in general. Sorry!!

THIRD UPDATE (16\January/2014): If one can prove that   $H=\langle g,H\cap M\rangle$ for all subgroups $H$ containing $g$, we are done. Anyway the gap in the first UPDATE is serious, and all my UPDATES do not say more than what has been said by Russ.

If $G$ is a non-trivial finite group which can be generated by two non-trivial elements of prime power orders, then the answer to the question is affirmative. Let $\mathcal{G}$ be the set of all subgroups of $G$ and let $G=\langle x,y\rangle$, where $x$ and $y$ are non-trivial elements of $G$ of prime power oders. Assume $A=\lbrace H\in \mathcal{G} \;|\; x\in H, y\not\in H \rbrace$, $B=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\in H \rbrace$, $C=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\not\in H \rbrace$ and $E=\lbrace H\in \mathcal{G} \;|\; x\in H, y\in H \rbrace$. So $E=\{G\}$ and $|C|\geq 1$ as $\{1\}\in C$. It follows that either $|A|\leq |\mathcal{G}|/2$ or $|B|\leq |\mathcal{G}|/2$.

I think (if I am not wrong) that every (if not "most" of!) finite simple groups can be generated by two elements of prime power orders. So the question has affirmative answer for finite simple groups and $2$-generated groups of prime power orders.

UPDATE (16/January/2014): Inspired by the Russ Answer, one can show that the question has affirmative answer for all finite groups $G$. Let $\Phi(G)$ be the Frattini subgroup of $G$ which by definition is the intersection of all maximal subgroups of $G$. Note that $G\not=\Phi(G)$, as $G$ is finite and it has at least one maximal subgroup. If $g$ is any prime power order element of $G\setminus \Phi(G)$, then $G=M\langle g\rangle$ for some maximal subgroup $M$ of $G$ so that $g\not\in M$. Now the map $H\mapsto H\cap M$ is injective for all subgroups $H$ containing $g$. So we may assume that all prime power order elements are in $\Phi(G)$ which implies that $G=\Phi(G)$, a conradiction, since a finite group can be generated by its Sylow subgroups.

SECOND UPDATE(16/January/2014) I found a gap in the above proof (Update), why $G=M\langle g\rangle$? It may not hold in general. So the answer of the question is still open for myself in general. Sorry!!

If $G$ is a non-trivial finite group which can be generated by two non-trivial elements of prime power orders, then the answer to the question is affirmative. Let $\mathcal{G}$ be the set of all subgroups of $G$ and let $G=\langle x,y\rangle$, where $x$ and $y$ are non-trivial elements of $G$ of prime power oders. Assume $A=\lbrace H\in \mathcal{G} \;|\; x\in H, y\not\in H \rbrace$, $B=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\in H \rbrace$, $C=\lbrace H\in \mathcal{G} \;|\; x\not\in H, y\not\in H \rbrace$ and $E=\lbrace H\in \mathcal{G} \;|\; x\in H, y\in H \rbrace$. So $E=\{G\}$ and $|C|\geq 1$ as $\{1\}\in C$. It follows that either $|A|\leq |\mathcal{G}|/2$ or $|B|\leq |\mathcal{G}|/2$.

I think (if I am not wrong) that every (if not "most" of!) finite simple groups can be generated by two elements of prime power orders. So the question has affirmative answer for finite simple groups and $2$-generated groups of prime power orders.

UPDATE (16/January/2014): Inspired by the Russ Answer, one can show that the question has affirmative answer for all finite groups $G$. Let $\Phi(G)$ be the Frattini subgroup of $G$ which by definition is the intersection of all maximal subgroups of $G$. Note that $G\not=\Phi(G)$, as $G$ is finite and it has at least one maximal subgroup. If $g$ is any prime power order element of $G\setminus \Phi(G)$, then $G=M\langle g\rangle$ for some maximal subgroup $M$ of $G$ so that $g\not\in M$. Now the map $H\mapsto H\cap M$ is injective for all subgroups $H$ containing $g$. So we may assume that all prime power order elements are in $\Phi(G)$ which implies that $G=\Phi(G)$, a conradiction, since a finite group can be generated by its Sylow subgroups.

SECOND UPDATE(16/January/2014) I found a gap in the above proof (Update), why $G=M\langle g\rangle$? It may not hold in general. So the answer of the question is still open for myself in general. Sorry!!

THIRD UPDATE (16\January/2014): We may fill the gap as follows:Since $G=\langle g, M\rangle$, it follows that $H=\langle g,H\cap M\rangle$, by Dedekind Modular Law. Now the reset of the proof is valid, I hope.