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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.

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    $\begingroup$ Recall that a CLT (Converse of Lagrange's Theorem) group is a group with the property that for any divisor of the order of the group there exists a subgroup of that order. Since it is known that finite abelian groups, as well as $p$-groups are CLT groups, this leaves a smaller class of subgroups to consider. $\endgroup$ – user1337 Apr 23 '15 at 13:16
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    $\begingroup$ A finite group $G$ of order $n$ such that the number of elements $x$ with $x^d=1$ is less than $d$ for all divisors $d$ of $n$ has to be cyclic. I wonder, if a variation of this topic can lead to a proof that there are at least as many subgroups as divisors. $\endgroup$ – j.p. Apr 23 '15 at 13:24
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    $\begingroup$ @j.p. That idea looks promising. One thing, should the phrase "is less than $d$" be replaced by "is at most $d$" in your comment? [though that would imply your version, I seem to recall one approach to doing the counting argument for showing some group is cyclic as involving the number being at most $d$] And +1 on the comment. $\endgroup$ – coffeemath Apr 23 '15 at 13:42
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    $\begingroup$ A short script in Maple shows that there are no such groups among the 793 groups of order $< 2^5 \cdot 3 = 96$. $\endgroup$ – Travis Willse Apr 23 '15 at 13:43
  • $\begingroup$ @coffeemath: Yes, you are right. It should have been "at most". $\endgroup$ – j.p. Apr 23 '15 at 14:04
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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.

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  • $\begingroup$ Derek-- Nice proof (seems to me). +1 on it for now, and likely I'll "accept" after I get a chance to go over it in detail. $\endgroup$ – coffeemath Apr 23 '15 at 14:46
  • $\begingroup$ How do you know that $P$ exist ? $\endgroup$ – Belgi Apr 23 '15 at 14:57
  • $\begingroup$ @Belgi $G$ has a $p$-Sylow subgroup $T$. Let's say $T$ is of size $p^k$; then $T$ will have a subgroup of order $p^i$ provided $i\leq k$. $\endgroup$ – mathmandan Apr 23 '15 at 15:06
  • $\begingroup$ @mathmandan - How do you know such subgroups of $T$ exist ? $\endgroup$ – Belgi Apr 23 '15 at 15:10
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    $\begingroup$ I was thinking of the proof as an induction on $m$ for all $n$ simultaneously. That is, when doing the proof for $m$, we assume that the result is true for all smaller $m$ and all $n$. We don't use induction on $n$ anywhere. I remember when thinking about this originally, the chief problem was to get an induction to work, and the key idea in the proof is actually to induct on $m$ rather than on $n$. $\endgroup$ – Derek Holt Apr 23 '15 at 17:06

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