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

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

• Do you assume that there exists a maximal subgroup? – j.p. Feb 13 '16 at 8:22
• @j.p. : Off-course ... when I say that every proper subgroup is contained in a maximal subgroup , it does mean that $\{e\}$ is also contained in a maximal subgroup , so yes there exist a maximal subgroup .. – user228168 Feb 13 '16 at 9:23
• @Arthur: see the Prüfer group for an example of a group without maximal subgroups. – Marc Paul Feb 13 '16 at 13:46
• @MarcPaul : Yes , a group might not have any maximal subgroup at all ( unlike rings with unity , by Zorn's Lemma ) ; that's why I have mentioned that every proper subgroup is contained in some maximal subgroup – user228168 Feb 13 '16 at 14:08
• Since there are maximal subgroups of finite index, there is a normal subgroup $N$ of finite index, and then $G/N = \langle xN \rangle$ is cyclic. A maximal subgroup containing $N$ cannot be conjugate to one containing $x$. So $G = \langle x \rangle$ is cyclic. – Derek Holt Feb 13 '16 at 14:34

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 $G = \langle a \rangle$.
• Why is the Frattini subgroup is the intersection of " finitely many " conjugates of $M$ ? I don't understand why are there only finitely many ... – user228168 Feb 13 '16 at 14:18
• The conjugates are of the form $g^{-1} M g = g^{-1} M \cdot M g$ (notice that $M\cdot M = M$ because $M$ is a subgroup), so in particular they are the product of a left coset with a right coset. Since there are only finitely many left cosets and finitely many right cosets, there can only be finitely many conjugates. – Marc Paul Feb 13 '16 at 14:30