$G$ be a group such that every maximal subgroup is of finite index and any two maximal subgroups are conjugate

Question

Let $G$ be a group such that every maximal subgroup is of finite index and any two maximal subgroups are conjugate and any proper subgroup is contained in a maximal subgroup . Then is $G$ cyclic ? I know that if $G$ is a finite group such that any two maximal subgroups are conjugate then $G$ is cyclic . But I cannot handle the infinite case . Please help . Thanks in advance

Answer 1

Here is a reduction to the finite case. Suppose $G$ is a group satisfying your requirements, and $M$ a maximal subgroup. Then all maximal subgroups are conjugate to $M$. This implies that the Frattini subgroup $\Phi(G)$ of $G$ (which is defined to be the intersection of all maximal subgroups of $G$) is the intersection of the finitely many conjugates of $M$, all of which have finite index, so $\Phi(G)$ has finite index. Therefore $G/\Phi(G)$ is finite.

Now it is easy to see that $G/\Phi(G)$ still has all maximal subgroups conjugate, so by the finite case, $G/\Phi(G)$ is cyclic. Let $\overline a$ be a generator of this group, where $a \in G$. Then we get that $G = \langle a, \Phi(G)\rangle$.

We now claim that $\langle a\rangle = G$. If not, then there is some maximal subgroup $N$ of $G$ containing $\langle a\rangle$. But by definition of the Frattini subgroup, $N$ also contains $\Phi(G)$, so then $N$ would contain $\langle a, \Phi(G)\rangle = G$. Contradiction, so there is no maximal subgroup containing $\langle a\rangle$, and therefore we conclude that $G = \langle a \rangle$.

Answer 2

Lemma 1: If $G$ is a finite group with the following properties, then G is cyclic.

• Every proper subgroup of $G$ is in one of its maximal subgroups
• Every Maximal subgroups of G are finite index.
• All maximal subgroups make one conjugacy class.

Proof: Assume that $M$ is a maximal subgroup in $G$. Then as $M$ is maximal, we should have $N_G(M)=M \;\text{or}\; N_G(M)=G$.

Case 1 $N_G(M)=G$: This means that $M$ is a normal subgroup in $G$; thus, the only maximal subgroup of $G$. This is because of the our third assumption. Now if $a \in G-M$ ( we have such an element because $M$ is a maximal subgroup of $G$): $$ \langle a \rangle \;\text{is a subgroup}\; G \;\text{but}\; \langle a \rangle \nsubseteq M$$ As a result, $\langle a \rangle$ could not be a proper subgroup, and this means that $\langle a \rangle =G$. This case is actually working when the group is not finite too.

Case 2 $N_G(M)=M$: We know the number (actually the cardinality) of elements in the conjugacy class of $M$ is $[G:N_G(M)]$. As a result: $$[G:N_G(M)]=[G:M]=n =\text{the number of maximal subgroups in the group}\; G$$ We can use $n$ in the above statement as we know the index of maximal subgroups of $G$ is finite. If the group $G$ is not cyclic then every element in $G$, such as $x$ should be in one of maximal ideals. Thus, by calculating the number of elements in all maximal ideals we should have $ |G| $ elements: $$ |G|-1 \leq (\text{number of elements in all maximal subgroups except the 1}) \leq (|M|-1)[G:N_G(M)] = (|M|-1)[G: M] \lt |G|-1$$ We use the fact that all conjugate subgroups have the same order in the above calculation. Therefore, the case 2 could not happen when $G$ is a finite group with the aforementioned properries.

Back to the proof of the question: let $N$ be the intersection of the all maximal ideals. We know that if $H$ a subgroup of group $G$, then the intersection of all its conjugates will be a normal subgroup in $G$, and this is the biggest normal subgroup of $G$ in $H$. This comes from the homomorphism that sends $G$ to $S_{[G:H]}$ because of the action of $G$ on the left cosets of $H$. The kernel of this homomorphism is the intersection of all conjugates of $H$. Therefore:

• $N$ is a normal subgroup of $G$.

• For each maximal subgroup of $G$, such as $M$, the subgroup $N$ is the biggest normal subgroup $G$ in $M$.

Lemma 2: $N$ has finite index in $G$.
Proof: it is true because $G/N$ is isomorphic with one of subgroups of group $S_{[G:M]}$. We know $[G:M]$ is finite so $G/N$ is finite too.

$G/N$ has all the three properties and it is finite from lemma 2, As a result of lemma 1, the group $G/N$ is cyclic. We know this group should be a $Z_p$, because it is a simple group. We cannot have a normal subgroup in $G$ which contains $N$ (except $N$ itself and $G$). Because every proper subgroup is in one the maximal subgroups, and $N$ is the biggest normal subgroup of $G$ in all maximal subgroups of $G$.

We assume $xN$ is the generator of $G/N$. Thus, $x^p \in N$. $$ G = (xN) \cup (x^2N) \cup \dots \cup (x^{p-1}N)$$

Lemma 3: $x$ cannot be in any of maximal subgroups.
Proof: if $x \in M$ and $M$ a maximal subgroup of $G$. we should have $x^i \in M$ for $i \in \{1,2, \dots p\}$. But, we have $N \subseteq M$. This means that: $$ G = \langle x, x^2, \dots x^{p-1},N \rangle \subseteq M$$

Thus, $x$ should not be in any maximal subgroups of group $G$, this means that the group generated by $x$ could not be a proper subgroup of G. $$ \langle x \rangle = G$$