Groups with "few" subgroups

If $G$ is a finite group of order $n,$ and the number of divisors of $n$ is $k,$ can $G$ have fewer than $k$ subgroups?

A cyclic group $G$ of order $n$ has exactly one subgroup for each divisor of $n$, so in this case $G$ has exactly $k$ subgroups.

For the alternating group $G=A_4,$ $n=12$ and $k=6.$ There is no subgroup of order $6$ in $A_4,$ yet there are ten subgroups, which exceeds $k=6$ for this case. I began to wonder if there could be a group for which so many divisors of the order of $G$ had no subgroups of that order, that there wound up being fewer subgroups in $G$ than the number of divisors of $|G|,$ hence my question.

Solution 1:

The answer to your question is no, that is not possible. This is a consequence of the following slightly more general result. Let $d(n)$ is the number of divisors of $n$.

Theorem Let $G$ be a finite group of order $n$. Then for any divisor $m$ of $n$, there exist at least $d(m)$ subgroups of $G$ of order dividing $m$.

Proof We prove the statement by induction on $m$, for all finite groups $G$. It's clear for $m=1$.

For $m>1$, let $p$ be a prime dividing $m$, and let $m = p^r k$ with $\gcd(p,k)=1$.

Since $d(m) = (r+1)d(k)$, it is sufficient to prove that, for each $i$ with $0 \le i \le r$, there are at least $d(k)$ subgroups of $G$ of order $p^i j$, for some $j$ that divides $k$.

So fix an $i$, let $P$ be any subgroup of $G$ of order $p^i$, let $N = N_G(P)$ be the normalizer of $P$ in $G$, and let $h = \gcd(|N|,k)$.

Since $h \le k < m$ and $h$ divides $|N/P|$, by inductive hypothesis, the number $s$ of subgroups $S/P$ of $N/P$ of order dividing $h$ satisfies $s \ge d(h)$. For each of these subgroups, the inverse image $S$ of $S/P$ in $N/P$ is a subgroup of $G$ of order $p^i j$, where $|S/P| = j$ divides $h$, and so $j$ divides $k$.

Now $P$ has $|G|/|N|$ distinct conjugates $P^g$ in $G$, and the $s|G|/|N|$ subgroups $S^g$ are all distinct. But $|G|/|N| \ge k/h$, so $s|G|/|N| \ge d(h)k/h \ge d(k)$, which completes the proof.

After thinking about it again, I think we can say that a group of order $n$ with exactly $d(n)$ subgroups must be cyclic. The above proof produces at least $d(n)$ subgroups with a normal $p$-subgroup for any prime divisor $p$ of $n$. So if there were exactly $d(n)$ subgroups, then all subgroups would have all of their Sylow subgroups normal, so they would all, including $G$ itself, be nilpotent. But a non-cyclic $p$-group $P$ has more than one subgroup of index $p$, and hence has more than $d(|P|)$ subgroups, so all Sylow subgroups of $G$ are cyclic and hence so is $G$.

By the way, I first posted this proof here in 2003. I would welcome any references to any other proofs.