Let $G$ be a finitely generated group and $G$ has exactly one maximal subgroup. Then I can conclude that $G\cong\mathbb Z_{p^k}$.

Now, I am looking for an example of infinite abelian group $G$ such that $G$ has exactly one maximal subgroup.

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    $\begingroup$ I do not understand "Then I can conclude that $\mathbb{Z}_{p^k}$."... $\endgroup$ – user87543 Dec 10 '13 at 6:44
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    $\begingroup$ @PraphullaKoushik If $G$ is not a cyclic group, then $a\in G-M$. Since $G$ is FG, $\langle a\rangle \subseteq M'$ where $M'\neq M$ is other maximal subgroup and It is a contradiction. $\endgroup$ – Babak Miraftab Dec 10 '13 at 6:47
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    $\begingroup$ I did not said what you said is correct or wrong... What i said is you have not said anything... Do you somehow mean "Then I can conclude that $G\cong \mathbb{Z_{p^k}}$"? $\endgroup$ – user87543 Dec 10 '13 at 7:03
  • $\begingroup$ @PraphullaKoushik: I think exactly as yours. $\endgroup$ – mrs Dec 10 '13 at 7:04
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    $\begingroup$ @PraphullaKoushik First I showed that $G$ is a cyclic group. Since $G$ has exactly one maximal subgroup, by The fundamental theorem of finitely generated abelian groups, I can conclude that $G\cong\mathbb Z_{p^k}$ $\endgroup$ – Babak Miraftab Dec 10 '13 at 12:48

Let $\mathbf Z(3)$ be the Prüfer 3-group. Then $$\mathbf Z/2\mathbf Z \times \mathbf Z(3)$$

has only one maximal subgroup, namely $0 \times \mathbf Z(3)$.

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    $\begingroup$ It might be nice to mention why there is only one maximal subgroup (the index must be 2 or 3, if it is 2 it must contain the Sylow 3, if it is 3, it must contain the Sylow 2, so look in Z(3) which has no maximal subgroups). $\endgroup$ – Jack Schmidt Dec 10 '13 at 17:39
  • $\begingroup$ @JackSchmidt the case index 2 is clear for me but it is not clear when $[G:N]=3$. I don't understand its relation with $Z(3)$ which has no maxmial subgroup. $\endgroup$ – Babak Miraftab Dec 12 '13 at 14:56
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    $\begingroup$ @Babgen: If [G:N] = 3, then N contains the Sylow 2-subgroup P (G/N contains no elements of order 2, so every Sylow 2-subgroup is contained in the kernel of G -> G/N). We consider the group G/P. N/P is still an index 3 subgroup, but G/P = Z(3) contains no index 3 subgroups [ every proper subgroup of Z(3) is finite, not finite index ]. $\endgroup$ – Jack Schmidt Dec 12 '13 at 14:59
  • $\begingroup$ @JackSchmidt thanks alot It is a genius solution. $\endgroup$ – Babak Miraftab Dec 12 '13 at 15:29

Looking at m_l's and Bruno's answers (and questions), one finds a reasonable classification of examples:

Theorem: An abelian group $G$ has exactly one maximal subgroup if and only if there is exactly one prime $p$ with $G \neq pG$ and for that prime $p$, $G/pG$ has order $p$.

Proof: If $M$ is a maximal subgroup of $G$, then $G/M$ is simple, so cyclic of order $q$ for some prime $q$. We count the maximal subgroups by prime $q$. For each prime $q$, $G/qG$ is a vector space of dimension say $d$ over $\mathbb{Z}/q\mathbb{Z}$ and so has $(q^d-1)/(q-1)$ maximal subspaces (or just $2^d$ if $d$ is infinite). Hence we must have $d=0$ for all but one prime, say $p$, where $d=1$. $\square$

$p$-group case

If $G$ is an abelian $p$-group, then $G/p^kG$ is a bounded abelian group, so a direct sum of cyclic $p$-groups. In particular, if $G/pG$ is cyclic, then $G/p^kG$ is also cyclic of order dividing $p^k$. Assume, by way of contradiction, that the size of the quotients are unbounded. Then clearly $G/p^kG \cong \mathbb{Z}/p^k\mathbb{Z}$. Now consider a collection of compatible epimorphisms from $G$ to $\mathbb{Z}/p^k\mathbb{Z}$. This gives a (nonzero) homomorphism from $G$ to the limit $\hat{\mathbb{Z}}_p$ with kernel $p^\infty = \cap_{k=1}^\infty p^k G$. However, $\hat{\mathbb{Z}}_p$ is torsion-free and so its kernel is all of $G$. That is, $p^\infty G=pG=G$, but this contradicts the homomorphism being nonzero.

Therefore, there is some largest quotient, $G/p^kG$ of size $p^k$, and $p^\infty G = p^{k+1}G = p^kG$. Let $G/p^kG = \langle x + p^k G \rangle$. Then $\langle x \rangle + p^i G = G$ for all $i$ (for $i \leq k$ by properties of cyclic groups, for $i \geq k$ because $p^i G = p^k G$). Hence $[ p^i G \cap \langle x \rangle : \langle x \rangle ] = [p^i G : \langle x \rangle + p^i G ] = [ p^i G : G ] = p^{\min(i,k)}$ since $G/p^i G$ is cyclic. However, $\langle x \rangle$ has a unique subgroup of that index, namely $p^i \langle x \rangle$. Hence $\langle x \rangle$ is a pure subgroup of $G$. Since it is also bounded, it is a direct summand and $G= \langle x \rangle \oplus B$. Now $pG = p \langle x \rangle \oplus pB$ has index $p$, so $pB=B$ is divisible. Thus $$G \cong \mathbb{Z}/p^k\mathbb{Z} \oplus \left( \mathbb{Z}[\tfrac1p]/\mathbb{Z}\right)^{(v_p)}$$ for some positive integer $k$ and some cardinal number $v_p$.

Torsion case

If we require $G$ to be torsion, then it is a direct sum of its Sylow $q$-subgroups, which must themselves be direct sums of Prüfer $q$-groups since they are $q$-divisible abelian $q$-groups. The Sylow $p$-subgroup must be a direct sum of $\mathbb{Z}/p^k\mathbb{Z}$ and a divisible $p$-group as above. Thus if $G$ is torsion, then $$G \cong \mathbb{Z}/p^k\mathbb{Z} \times \left( \bigoplus_{q} \left(\left.\mathbb{Z}\left[\tfrac1q\right]\middle/\mathbb{Z}\right.\right)^{(v_q)} \right)$$ for some positive integer $k$, and some cardinal numbers $v_q$ for each prime $q$, including $q=p$.

Bruno took $p=2$, $k=1$, $v_3=1$, and $v_q =0$ for $q\neq 3$. Thanks to babgen for pointing out several problems in earlier versions.

Rank 1 case

Torsion-free rank 1 groups (subgroups of $\mathbb{Q}$) are described up to isomorphism by their “type” which basically says how many times the average element is divisible by each prime. That is, for a type $t$, $A_t = \left\{ \tfrac ab \in \mathbb{Q} : v_p(b) \leq t_p, \forall p \right\}$, but two types $s,t$ are considered equivalent if they differ finitely, $\sum_p \left|t_p-s_p\right| < \infty$.

In our case, we have $t_q = \infty$ for all primes except one, $p$, and by the definition of types it doesn't matter what we set $t_p$ (@m_l's answer sets it to “$k-1$”). I'll choose $t_p=0$ for simplicity.

$$A = \left\{ \frac{a}{b} : b \text{ is not divisible by } p \right\}$$

Then $qA = A$ for all primes $q \neq p$, but $A/pA$ is order $p$, generated by $1+pA$.

These are the only torsion-free rank 1 examples up to isomorphism (one isomorphism class for each prime $p$).


A fancier version of $A$ occurred to me first, and I record it here for what it is worth:

Let $J$ be the additive group of $p$-adic integers. The subgroup $pJ$ is a maximal subgroup, since it has index $p$.

Let $M$ be a maximal subgroup of $J$. We want to show $M=pJ$.

Assume, by way of contradiction, that the index of $M$ in $J$ is not $p$, but rather some other prime number $q$. Let $x \in J$. Since $q$ is relatively prime to $p$, the $p$-adic expansion of $\tfrac1q$ lies in $J$, so that $x = qy$ for some $y \in J$ (namely, the $p$-adic expansion of $y=\tfrac xq$). However, by Lagrange's theorem, $qJ \leq M$ so that $x = qy \in M$, and $J=M$.

Hence the index of $M$ in $J$ is $p$, so Lagrange shows again that $pJ \leq M$, but $pJ$ is maximal, so $pJ=M$.

  • $\begingroup$ Very nice and complete answer. I enjoyed reading it. :) $\endgroup$ – m_l Dec 11 '13 at 0:02
  • $\begingroup$ @JackSchmidt I can't apply your strategy for an abelian q-group. can you more explain when G is an abelian q-group? $\endgroup$ – Babak Miraftab Dec 12 '13 at 17:11
  • $\begingroup$ @JackSchmidt I guess that abelian p-group with exactly one maximal is one of the following groups $\mathbb Z_{p^k}\times\mathbb Z_{p^{\infty}}$ or $\mathbb Z_{p^k}$. $\endgroup$ – Babak Miraftab Dec 12 '13 at 17:24
  • $\begingroup$ @Babgen: thanks, all fixed I hope. $\endgroup$ – Jack Schmidt Dec 18 '13 at 21:05
  • $\begingroup$ Dear jack What did you mean about $\hat{\mathbb{Z}}_p$ exactly? and can you more explain why kernel is all of $G$ $\endgroup$ – Babak Miraftab Dec 23 '13 at 19:07

For any prime $p$ and natural number $k$, the additive group of the localisation $$\mathbb{Z}_{(p^k)} = \left\{ \frac{a}{b} \in \mathbb{Q} : \text{gcd}(a,b) = 1,~p^k \not\mid b \right\}$$ is an abelian group with exactly one maximal subgroup: $p\mathbb{Z}_{(p^k)}$.

  • $\begingroup$ @m_1 Why is every subgroup contained in a maximal subgroup? for example $\mathbb Z_{p^k}$. $\endgroup$ – Babak Miraftab Dec 10 '13 at 12:43
  • $\begingroup$ @Babgen: Yeah, I had a feeling there had to be a mistake in there. You are right, my point does not hold. I should get more sleep. Will try to fix it, if I can't do so I will delete this answer. I don't see what you are getting at with $\mathbb{Z}_{p^k},$ though. For finite groups, clearly every subgroup is contained in a maximal subgroup. $\endgroup$ – m_l Dec 10 '13 at 16:38

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