Determining subgroups whose index is congruent to its order modulo n.
Consider a finite group $G$ and a subgroup $H$ of $G$. We are interested in determining the $n \in \mathbb{N}$ for which $[G : H] \equiv |H|\pmod{n}$. That is, we wish to determine when the index of a subgroup is congruent to its order modulo n, or when $$n \mid [G : H] - |H|$$ We want to determine how many such subgroups a group $G$ will have for a given $n$, and how this relates to the $|G|$ and $|H|$. We will refer to these subgroups as n-congruent subgroups. We will also look at how these groups are related to Hall subgroups.
$H$ is an n-congruent subgroup of $G$ if $H \le G$ and $[G : H] \equiv |H|\pmod{n}$.
By Lagrange's Theorem, we can write $|G| = |H|[G : H]$, so the condition of an n-subgroup puts an arithmetic constraint on the factorization of $|G|$.
If $H$ is an n-congruent subgroup of $G$ then $|G| \equiv |H|^2 \pmod{n|H|}$
Let $ m = |H|, k=[G:H]. $ Then \(mk=|G|\), and the condition for n-congruence becomes $$ k \equiv m \pmod n. $$ Or equivalently, $$ n \mid (k-m). $$ Since \(k=|G|/m\), this may be rewritten as $$ n \mid \left(\frac{|G|}{m}-m\right). $$ Multiplying by \(m\), $$ nm \mid (|G|-m^2). $$ So every \(n\)-congruent subgroup gives a divisor \(m\) of \(|G|\) satisfying $$ |G| \equiv m^2 \pmod{nm}. $$ $ \blacksquare $
If $H$ is an n-congruent subgroup of $G$ then $|H|^2 \equiv |G| \pmod{n}$
Using the same notation as the previous theorem, the congruence $$ k\equiv m \pmod n $$ implies $$ mk \equiv m^2 \pmod n. $$ Since \(mk=|G|\), $$ |G| \equiv m^2 \pmod n. $$ $ \blacksquare $
This gives a necessary condition. In other words, the order of \(H\) must be a square root of \(|G|\) modulo \(n\). This connects the problem to quadratic congruences.
We now look at the interesting case when $n=|G|$. The following theorem allows us to quickly determine if $H$ is $|G|$-congruent.
$H$ is \(|G|\)-congruent if and only if \(H\) has order \(\sqrt{|G|}\).
If $H=G$ then the condition for $|G|$-congruence becomes $[G : G] \equiv |G|\pmod{|G|}$, so $1 \equiv |G|\pmod{|G|}$, which only holds when $G$ is trivial. So we only consider the case when $H$ is a proper subgroup of $G$.
$(\implies)$ For the forward case, suppose $H$ is $|G|$-congruent. Then, $$ [G:H]\equiv |H|\pmod{|G|}. $$ Since both numbers are positive and less than \(|G|\), the congruence forces $$ [G:H]=|H|. $$ Hence $$ |G|=|H|^2. $$ And, $$ |H| = \sqrt{|G|} $$
$(\impliedby)$ Suppose $|H| = \sqrt{|G|}$. To show that $H$ is $|G|$-congruent we need to show that $|G| \mid [G:H] - |H|$, or equivalently $|G| \mid \frac{|G|}{\sqrt{|G|}} - \sqrt{|G|}$. Multiplying by $\sqrt{|G|}$, gives $\sqrt{|G|}|G| \mid |G| - |G| = 0$, which is trivially true. $ \blacksquare $
Groups of square order are particularly relevant to this case.
$Z_{36}$ contains a subgroup of order 6, so by the above theorem, it has a $|G|$-congrugent subgroup. The same is true for any finite cyclic group of square order.
$S_{3}$, the symmetric group of order 6, does not have a $|G|$-congrugent subgroup, since $\sqrt{6}$ is not an integer and so $G$ has no subgroup of this order.
Recall that \(H\) is a Hall subgroup if $$ (|H|,[G:H])=1. $$ Hall subgroups and $|G|$-congruent subgroups interact interestingly. Suppose \(H\) is Hall and \(n=|G|\). Then \(H\) is \(n\)-congruent exactly when $$ |H|=[G:H]. $$ Combined with coprimality, this forces $$ |H|=[G:H]=1. $$ So Hall subgroups and \(|G|\)-congruent subgroups are essentially incompatible except in trivial cases. This leads to the following theorem.
$H$ is a proper $|G|$-congruent subgroup of $G$ if and only if H is not a Hall subgroup of G.
If $H$ is a proper $|G|$-congruent subgroup of $G$, then $[G:H]=|H|$. If $H$ was also a Hall subgroup, then both $H$ and $G$ would be the trivial group, contradicting the fact that $H$ is a proper subgroup.
Conversely, if $H$ is a Hall subgroup of $G$, $(|H|,[G:H])=1$. If $H$ was also a proper $|G|$-congruent subgroup, then again this leads to the contradition that both $G$ and $H$ are the trivial group. $ \blacksquare $
The above theorem allows us to save a couple of computations for showing a given subgroup is not Hall.
Let $H$ be the subgroup of order 88 of $G = Z_{7744}$, so $|G|=88^2$. Then, $[G:H]=|H|$ and $H$ is a proper $|G|$-congruent subgroup, thus H is not a Hall subgroup.
We've introduced the concept of an n-congrugent subgroup and shown some basic properties of them. The natural next question to explore is the existence question, namely for which pairs \((G,n)\) does an \(n\)-congruent subgroup exist? The classification of the n-congruent subgroups of cylic and finite Abelian groups should provide valuable insight into this question. We will explore these questions in further work.
[2] Dummitt, D.S. and Foote, R.M. (2004) Abstract Algebra. 3rd Edition, John Wiley & Sons, Inc.
[3] Gallian, J. (2021). Contemporary Abstract Algebra (10th ed.). Chapman and Hall/CRC