0
$\begingroup$

Under what conditions can it be determined whether or not any group G has some subgroup N such that the quotient group G/N is a cyclic group? In other words, how can it be determined whether or not a group has a non-trivial cyclic quotient group?

I know that for any group G, if there is a group N of order g/p where g is the order of G and p is the smallest prime factor of the order of that group, then the group N will be normal and G/N will be a prime order cyclic group. Does the group N always exist for a finite group? Under what conditions does it exist?

What about for infinite groups? In what cases can we say they have a subgroup N such that G/N is a cyclic group?

My motivation in asking this question is that for any cellular automaton C on a group G, C can simulate a simpler cellular automaton if every element in each coset of G is the same. For example, if G is the group of a two dimensional grid, which corresponds to the free abelian group on two generators, making every element in each column have the same state causes the columns, which are cosets of G, to act as a one dimensional cellular automaton. The group of this "factor" cellular automaton is N, the free abelian group on one generator.

I am curious to know under what conditions any group could simulate a 1-dimensional cellular automaton, hence, under what conditions it has a cyclic quotient group.

$\endgroup$
9
  • 1
    $\begingroup$ This question would benefit from added context, as it is slightly odd and confused. (You seem to think maximal subgroups are normal?) For example, simply telling us where you got the problem from or explaining what you tried would be a big step forward! (For further feedback/help with asking questions, you can ask here.) $\endgroup$
    – user1729
    Aug 31, 2021 at 10:59
  • 2
    $\begingroup$ There are finite simple groups which are not cyclic like $A_5$. You need them to be Abelian for your claim to work. $\endgroup$ Aug 31, 2021 at 11:00
  • 3
    $\begingroup$ ...and there are infinite non-abelian simple groups. For instance, the alternating group of the positive integers (permutations with finite support and signature 1). $\endgroup$ Aug 31, 2021 at 11:03
  • 1
    $\begingroup$ While every subgroup of prime index is maximal, not every maximal subgroup is of prime index; and not every maximal subgroup is normal. There are finite groups that have no nontrivial cyclic quotient. $\endgroup$ Aug 31, 2021 at 16:08
  • 1
    $\begingroup$ If there is a subgroup whose index is the smallest prime that divides the order of the group, then that subgroup is normal. However, not all maximal subgroups have orders like that, and not every finite group has a subgroup whose index is the smallest prime that divides the order of the group: $A_5$ has order $60$, but no subgroup of order $30$ (index $2$, the smallest prime that divides $60$). And $A_5$ has a maximal subgroup of order $6$ (index $10$). $\endgroup$ Sep 1, 2021 at 0:42

1 Answer 1

3
$\begingroup$

Both the question and your answer contain several misconceptions/errors.

  1. Your question seems to assert that every finite group has a maximal normal subgroup of prime index. While every subgroup of prime index is necessarily maximal (this follows from the multiplicativity of the index, if $H\lt K \lt G$ then $[G:H]=[G:K][K:H]$), it is not true that every such group is necessarily normal, and it is not true that every maximal subgroup has prime index. For example, $A_5$ has a maximal subgroup of index $5$ (the subgroup fixing $5$, say), but it is not normal; and it has a maximal subgroup of order $6$ and index $10$, generated by $(123)$ and $(12)(45)$.

  2. The statement "if you divide the order of a finite group by the smallest prime possible, that any subgroup will be normal and of prime cyclic order" is hopelessly garbled. On the basis of your comment on the post, you tried to say that if the index of a subgroup is the smallest prime that divides the order of the group then that subgroup will be normal and the quotient cyclic of prime order. That is true, but not every finite group has such a subgroup (see above: $A_5$ has no subgroup of index $2$).

  3. The statement in the answer that "any group not composed of cyclic simple groups will not have any cyclic simple quotient" is false if you mean "a group in which not every composition factor is a cyclic simple group". It is false for $\mathbb{Z}$, for example; and for finite groups, it is false for $S_5$ (which has a quotient isomorphic to $C_2$) and more generally, any group of the form $A\times B$ where $A$ is finite abelian and $B$ is nonabelian simple. If you mean "for which every composition factor is not cyclic simple group", then that is true for finite groups but your phrasing is extremely confusing.

To your questions:

Theorem. A finite group $G$ has a quotient that is nontrivial cyclic if and only if the commutator subgroup is a proper subgroup, $G\neq [G,G]$.

Proof. A nontrivial finite abelian group has nontrivial cyclic quotients, hence nontrivial cyclic quotients of prime order. If $G=[G,G]$, then the only abelian quotient of $G$ is trivial so $G$ cannot have nontrivial cyclic quotients. If $G\neq[G,G]$, then $G/[G,G]$ is finite abelian and hence has itself a quotient which is cyclic of prime order, and therefore so does $G$ by the isomorphism theorems. $\Box$

For infinite groups the question is harder. There are groups (even abelian groups) with no subgroups of finite index. For example, the Prüfer $p$-group $\mathbb{Z}_{p^{\infty}}$ is infinite, but all proper subgroups are finite. And the additive group of rationals $\mathbb{Q}$ does not have any maximal subgroups, and in particular cannot have a subgroup of finite index. The quotients of $\mathbb{Z}_{p^{\infty}}$ are either trivial or isomorphic to $\mathbb{Z}_{p^{\infty}}$ itself, which is quasicyclic but not cyclic. No nontrivial quotient of $\mathbb{Q}$ can be nontrivial and cyclic (even infinite cyclic) because $\mathbb{Q}$ is divisible, hence so is every quotient, and no nontrivial cyclic group is divisible. Both of these groups are abelian, so the commutator test certainly does not work in general.

It does work for finitely generated groups:

Theorem. Let $G$ be a finitely generated group. Then $G$ has a quotient that is nontrivial cyclic if and only if $[G,G]\neq G$.

Proof. It follows as above once you verify that a finitely generated abelian group has such quotients, and note that a quotient of a finitely generated group is finitely generated. $\Box$

$\endgroup$
1
  • $\begingroup$ On point 3 I did mean the latter, but it definitely was confusing. I didn't really ask the question well but you answered it well anyways. $\endgroup$ Sep 1, 2021 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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