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I'm looking for a group for which the number of subgroups is more than the number of elements in the group! I tried a few possibilities - it can't be cyclic, and I think we'll have to consider group of infinite order.

senshin
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Mathronaut
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    The smallest example has order $4$. – Mikko Korhonen May 25 '14 at 15:16
  • @Yves: my apologies; I misspoke. – Qiaochu Yuan May 25 '14 at 20:01
  • To bounce on Qiaochu's erased comment, it's not hard to see that if $x=4^n$ (to simplify), and $G$ is a vector space over $F_2$ of dimension $2n$, then $G$ has at most $n2^{4n^2}=n2^{4\log_2(x)^2}$ subgroups, and has at least $2^{n^2}$ subgroups (namely graphs of linear maps between two spaces of dimension $n$); I'm lazy to check the precise behavior but it's essentially in $\exp(\log(x)^2)$. It would sound plausible that it's essentially the largest possible number of subgroups...? – YCor May 25 '14 at 20:18
  • concerning infinite groups, plenty of countable groups (e.g. a free group on 2 generators, or $\mathbf{Q}$) have $2^{\aleph_0}$ many groups. Also I guess that "most" uncountable groups $G$ have $2^{|G|}$ subgroups). – YCor May 25 '14 at 20:39
  • To @YCor's comment, see https://math.stackexchange.com/questions/257175/countable-group-uncountably-many-distinct-subgroup – Nate Eldredge Apr 19 '20 at 04:15

6 Answers6

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Consider the product of $n \gt 2$ copies of $\mathbb{Z}_2$, a group of order $2^n$. Each nonzero element in this (additive) group has order two, so in addition to the trivial subgroup, there are $2^n - 1$ subgroups of order two.

Of course there are also proper subgroups of order greater than two, so more subgroups than elements.

hardmath
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    $n = 2$ works also. – Mikko Korhonen May 25 '14 at 15:14
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    @MikkoKorhonen: Sort of. For $n=2$ we have the trivial subgroup and three subgroups of order two, for a total of exactly four proper subgroups. If you are allowed to count the group itself, that would give five "subgroups", but in the more restricted sense of proper subgroups we need $n \gt 2$. – hardmath May 25 '14 at 15:23
  • Sir, as you had written "There are proper subgroups of order greater than $2$" I know all they isomorphic to $C_2\times C_2$(since there is no element of order 4). But exactly how many? Please help – Akash Patalwanshi Mar 06 '20 at 05:51
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    @Akash: That's a good problem to ask as a separate Question. At first I thought I could respond with just a Comment, but on reflection I find that it needs more space. – hardmath Mar 06 '20 at 19:48
  • @bof: That's plausible. I think many of the arguments described in counting subgroups of $C_2^3$ will directly apply to $C_2^n$. In particular the subgroups of order $C_2^k$ are in one-to-one correspondence with those of order $C_2^{n-k}$ for $0\le k\le n$. See as well this closely related Answer. – hardmath Oct 20 '21 at 15:28
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    @Akash: I see that user joriki gave you a succinct solution to counting the subgroups of order four in $C_2^n$. It would have been helpful to link to this earlier Question for context. – hardmath Oct 20 '21 at 15:52
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$C_2 \times C_2 \times C_2$ has 8 elements. Each of the 7 non-identity elements generates a subgroup of order 2. Any pair of non identity elements also generates a subgroup isomorphic to $C_2 \times C_2$. There are 7 of these subgroups, for a total of 14 nontrivial proper subgroups.

Josh B.
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Number of subgroups of $D_{2n}=\sigma(n)+\tau(n)$, where $\sigma$ and $\tau $ are sum of divisors and number of divisors of $n$, so determine all $n$ for which $\sigma(n)+\tau(n)>2n$ and you will have alot many examples (just do not try primes bigger than $3$)

For $n=4$, $\sigma(n)+\tau(n)=10>4$, so $D_4 $ is an example.

For $n=5$, $\sigma(n)+\tau(n)$, not true.

For $n=6$, $\sigma(n)+\tau(n)=16>12$. so $D_6 $ is an example..

and so on....

Bhaskar Vashishth
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  • Sir please can you help me in this. Numbers of subgroups isomorphic to $\mathbb{Z_2}\times \mathbb{Z_2}$ (klien four group) in direct product of n-copies of $\mathbb{Z_2}$ – Akash Patalwanshi Mar 06 '20 at 05:57
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As alluded to in the comment by Mikko Korhonen,

the smallest example is the Klein Vierergruppe $V=\{1,a,b,c\}$, where $a^2=b^2=c^2=1$.

It has four elements and five subgroups: $\{1\}, V, \{1,a\}, \{1,b\}, $ and $\{1,c\}$.

J. W. Tanner
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Every uncountable abelian group has more subgroups than elements. First, suppose $G$ is an uncountable abelian group which is a direct sum of cyclic groups. The number of summands must be equal to $|G|$, and every subset of the summands generates a different subgroup, so $G$ has $2^{|G|}$ different subgroups.

Now suppose $G$ is an arbitrary uncountable abelian group. By the previous argument, it suffices to show $G$ has a subgroup of the same cardinality that is a direct sum of cyclic groups. Suppose $H\subseteq G$ is any subgroup such that $|H|<|G|$. We can enlarge $H$ to a subgroup $H'$ with $|H'|\leq |H|+\aleph_0$ such that if $h\in H'$, $n\in\mathbb{Z}$, and $h$ is divisible by $n$ in $G$, then $h$ is divisible by $n$ in $H'$ (just iterate adding witnesses to all such divisibilities to the group $\omega$ times). In particular, since $G$ is uncountable, we still have $|H'|<|G|$, so we can pick some $g\in G\setminus H'$. Let $n\in\mathbb{Z}_+$ be minimal such that $ng\in H'$ (or if no such $n$ exists, let $C$ be the subgroup generated by $g$ and jump to the next paragraph). Since $ng$ is divisible by $n$ in $G$, it is also divisible by $n$ in $H'$, so there is some $h\in H'$ such that $nh=ng$. By minimality of $n$, $g-h$ has order $n$, and the cyclic group $C$ that it generates has trivial intersection with $H'$.

We have thus shown that if $H\subset G$ is a subgroup with $|H|<|G|$, then there is a nontrivial cyclic subgroup $C\subseteq G$ such that $C\cap H=0$. If $H$ is a direct sum of cyclic groups, then $C+H$ will be as well, with $H$ as a direct summand. Iterating this transfinitely, we can keep finding larger and larger direct sums of cyclic groups which are subgroups of $G$, until we reach one of the same cardinality as $G$.

(Incidentally, this is not true for nonabelian groups, or at least it cannot be proved in ZFC: according to this answer, for instance, it follows from CH that there is a group of cardinality $\aleph_1$ with only $\aleph_1$ subgroups. I don't know whether a nonabelian counterexample can be proved to exist in ZFC.)

Eric Wofsey
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Intuitively, I would think there would be a pretty good chance. Considering all the ($2^n\gt n$) subsets.

Granted this is kind of naive, not all subsets are subgroups. Nevertheless, we have seen examples, both finite and infinite, where alot are.

And as the answers above show, including elementary abelian groups (the Klein four group included) and dihedral groups, the answer is yes.