Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The alternating group on five elements, $A_5$, has $1$ element order $1$, $15$ elements order $2$, $20$ elements order $3$, and $24$ elements order $5$. If I take these numbers mod some $n$, sometimes I get numbers of elements which correspond to a subgroup of $A_5$.

To clarify, an example with $n=10$. $\{1,15,20,24\}$ mod $10$ becomes $\{1,5,0,4\}$. $D_{10}$, the dihedral group of order $10$, has one element order $1$, $5$ elements order $2$, and $4$ elements of order $5$. $D_{10}$ is also a subgroup of $A_5$. This also happens for the other $n$ equal to the order of some subgroup of $A_5$.

My question is, why does this happen? And what conditions are necessary for this to occur in other groups, since this doesn't occur for all groups.

share|improve this question
I feel like it's just a coincidence. If you have more nasty examples though, I'd be willing to believe there might be something deep there. –  Patrick Da Silva Jan 16 '12 at 5:24
It's worth noting that if the property holds for a group $G$ it holds for all subgroups of $G,$ and that the property is not held by non-simple cyclic subgroups of order greater than 4. –  jspecter Jan 16 '12 at 5:33
On the otherhand, the property holds for all non-abelain groups of order $pq$ and elementary $p$-groups. –  jspecter Jan 16 '12 at 5:59
In $A_6$ there are {1,45,80,90,144} elements of orders {1,2,3,4,5} This property holds for n={1,2,3,4,5,6,8,9,10,18,36}. For n=12, we get 24 elements, $S_4$, and for n=30 we get 60 elements, $A_5$. –  Josh B. Jan 16 '12 at 11:38
@jspecter: I wrote up the consequences of your comments. it does classify nilpotent groups and simple groups. I'm hoping the generalization of the pq groups I mentioned is basically it for solvable groups, but I haven't worked with solvable groups recently enough to remember the techniques. Solvable groups definitely have lots of inconvenient subgroup orders, so I assume they are limited, but the exponent p p-groups combined with (actually existing) weird actions was too much for me today. –  Jack Schmidt Jan 16 '12 at 18:56

1 Answer 1

up vote 9 down vote accepted

Let $\omega_d(G)$ be the number of elements of order d in the finite group G. Let X be the collection of all finite groups G such that for every d, $\omega_d(H) = \omega_d(G) \mod |H|$ for every subgroup HG.

As jspecter's remarks, X is clearly closed under subgroups, so we may be interested in minimal non-X-groups, those groups which are not in X but every proper subgroup is in X.

Lemma: The cyclic group of order pq for distinct primes p, q is a minimal non-X-group.

Proof: It has p−1 elements of order p, but it cannot be the case that both p−1≡0 mod q and q−1≡0 mod p. $\square$

Lemma: A finite group with both a cyclic subgroup and an elementary abelian subgroup, both of order $p^2$, cannot be an X-group.

Proof: The number of elements of order p in the whole group must be equal to $p^2-1 \mod p^2$ due to the elementary abelian subgroup, but it must also be equal to $p-1 \mod p^2$ due to the cyclic subgroup. $\square$

Proposition: A finite abelian group is an X-group if and only if it is a cyclic p-group or an elementary abelian p-group for some prime p.

Proof: If the group is not a p-group, then the first lemma applies. If the group is not of exponent p, then it must be cyclic by the second lemma. $\square$

Proposition: A finite 2-group is an X-group if and only if it is cyclic, elementary abelian, or the quaternion group of order 8.

Proof: If the group has exponent 2, then it is elementary abelian. Otherwise it contains an element of order 4. By the second lemma it contains no elementary abelian subgroups of rank 2, and so it must be cyclic or generalized quaternion. However the generalized quaternion group of order 16 is a minimal non-X-group, and so the only possibility is Q8 which is easily checked to be an X-group, have 6 elements of order 4, which is 2 mod 4 and 0 mod 2. $\square$

The p-groups for odd p will defy such an explicit classification, since all groups of exponent p are X-groups. However, this is the only obstacle:

Proposition: A p-group for odd p is an X-group if and only if it is cyclic or of exponent p.

Proof: Again, unless it is exponent p it cannot have any elementary abelian subgroups of rank 2. In the odd p case, this leaves only the cyclic groups as candidates. $\square$.

Proposition: A nilpotent group is an X-group if and only if it is a cyclic p-group, a group of exponent p, or the quaternion group Q8.

Proof: A nilpotent X-group must be a p-group lest it contain a cyclic subgroup of order pq. $\square$

I will leave solvable groups open for now, but remark with jspecter that non-abelian groups of order pq and more generally cyclic groups of order $p^n$ acting faithfully and irreducibly on groups of order $q^m$ are X-groups, as they have $(p^d-p^{d-1})q^m$ elements of order $p^d$ which is 0 mod $q^m$ as it should be, and since $1 \equiv q^m \mod p^n$ in order for the action to exist, it is even equal to $p^d - p^{d-1}$ mod $p^d$ as it should be. Similarly, $q^m-1$ is equal to 0 mod $p^n$ and $q^d-1$ mod $q^d$, all as it should be.

Proposition: A finite non-abelian simple group is an X-group if and only if it is PSL(2,4) or PSL(2,8).

Proof: We have several convenient hypotheses available: the Sylow 2-subgroups are elementary abelian (cyclic and quaternion being ruled out by Glauberman's Z*-theorem), and so the group is either a $\operatorname{PSL}(2,2^n)$ or J1. Explicit calculation rules out J1, and confirms PSL(2,4) and PSL(2,8). For general $2^n$, our group contains cyclic groups of order $2^n+1$ and $2^n-1$, both of which must be prime powers. I believe the only solutions to this are $2^2\pm1=3,5$ and $2^3\pm1=7,9$, but I am not sure if this a well-known fact or an open problem in elementary number theory.

This can be extended to all almost simple groups by observing that S5 and PΓL(2,8) are not X-groups. The direct product of almost simple groups is never an X-group, since it contains a cyclic subgroup of order $2p$.

share|improve this answer
Thank you very much for your detailed answer. This is very helpful. –  Josh B. Jan 16 '12 at 21:28
@JoshB: No problem. Let me know if you get a good classification for solvable groups. If the Fitting is cyclic, or even abelian, I think I have a good idea of what the group looks like, but general exponent p confuses me. I suspect the non-solvable case will be easy once the solvable case is handled (or at least after the p-constrained case), since the generalized Fitting is so restricted. –  Jack Schmidt Jan 16 '12 at 23:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.