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I just want to make sure that every group $\mathbb{Z}_n$ for any $n$ is cyclic. Further, every group of prime order is cyclic because it is isomorphic to $\mathbb{Z}_p$. I think I have a handle on that, but what other finite groups are there that aren't cyclic? In particular, I'm looking for examples that aren't direct products of groups, just simply groups like $\mathbb{Z}_n$, not $\mathbb{Z}_a \times \mathbb{Z}_b$, etc.

David
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  • check out the so-called "klein group" of four elements all satisfying $g^2=e$. i'm not sure if the name comes from Felix Klein, or from the German word for "small" – David Holden Dec 11 '13 at 08:36
  • Dihedral groups, symmetric groups, alternating groups, groups of matrices over finite fields. – Gerry Myerson Dec 11 '13 at 08:37
  • @David, that's a direct product. – Gerry Myerson Dec 11 '13 at 08:37
  • Hear, hear! Hear, hear! (Redundancy supplies 15 characters!) – Robert Lewis Dec 11 '13 at 08:38
  • you are right, Gerry. as often i "didn't read the question thoroughly". apologies. – David Holden Dec 11 '13 at 08:39
  • @GerryMyerson, having gone over symmetric groups, I believe the symmetric group $S_4$, for example, is the group of permutations of 4 letters, correct? Are the subgroups of symmetric groups cyclic? – David Dec 11 '13 at 08:40
  • Yes, $S_n$ is the group of permutations of $n$ letters. Some of its subgroups are cyclic, some are not. E.g., the subgroups of $S_4$ include $S_3$, and the dihedral group of 8 elements, and the alternating group $A_4$, and those groups aren't cyclic. – Gerry Myerson Dec 11 '13 at 08:42
  • @GerryMyerson, I'm a bit perplexed by that. $S_4$ has $4!$ permutations, which could be expressed as cycles or the product of cycles. Aren't the cycles cyclic? – David Dec 11 '13 at 09:01
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    Well, each element of $S_4$ (or of any other group) generates a cyclic subgroup. $S_4$ has subgroups isomorphic to the cyclic group of order 1, of order 2, of order 3, and of order 4. But it also has those non-cyclic, non-commutative subgroups as well. – Gerry Myerson Dec 11 '13 at 09:07
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    Aha, @GerryMyerson, I think I'm getting it now. Everything became far too meta for me, but now I understand (I think). A cyclic group means the group can be generated by one element and powers of said element. Clearly, this isn't the case for all finite groups. For example, is it wrong to say that for $S_4$, the permutation given by (1,2)(3,4) is not cyclic? We can't generate the entire thing by starting from any of 1, 2, 3, 4. For example, starting at 1, we can't make our way back to 1 while simultaneously landing on 2, 3, 4. Now, (1,2) is cyclic, but (1,2)(3,4) is not. Am I correct? – David Dec 11 '13 at 09:10
  • I think you're mixing up "cycle" and "cyclic". It makes sense to ask whether an individual permutation is a cycle; $(1,2)$ is, $(1,2)(3,4)$ isn't. It doesn't make sense to ask whether an individual permutation is cyclic --- that's a property a group may or may not have. The group generated by $(1,2)(3,4)$ is cyclic. – Gerry Myerson Dec 11 '13 at 09:39

2 Answers2

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The easiest examples are abelian groups, which are direct products of cyclic groups. The Klein V group is the easiest example. It has order $4$ and is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$. As it turns out, there is a good description of finite abelian groups which totally classifies them by looking at the prime factorization of their orders.

The very first nonabelian group you run into will usually be dihedral groups, the symmetries of $n$-gons. These are just cyclic groups that can be "flipped" by a certain element called a reflection.

Next step up is symmetric groups, the group of bijective functions from $\{1,\ldots,n\}$ to itself (also known as permutations). Cayley's theorem shows how these relate to finite groups in general. (This is not a very useful result, though, for practical purposes, as permutation representations of groups are often the worst possible way to understand the group's structure.) Alternating groups are related to symmetric groups, too, but they're a bit harder to understand from a structural standpoint.

And if you really want to see a lot of examples of groups, you can read my answer here, though a lot of it may be out of your reach at this point.

Community
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Alexander Gruber
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  • link to wikipedia page? https://en.wikipedia.org/wiki/Klein_four-group as their explanation (and the Cayley table really helps!) – Jason S Jun 22 '17 at 14:59
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    Is "Klein V group" a typo? (I'm not sure what you were going for if it is a typo though. "Klein four-group $V$"? Although I have only previously seen $V_4$ used, not $V$, until I looked at the wiki article just now. Possibly I have seen $V$ used before and not clocked it though.) – user1729 Jan 25 '19 at 11:41
  • The Klein $V$ group is called such because the German term for the number 4 is "vier", – Harry Evans Sep 25 '19 at 04:22
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Every finite group is isomorphic to a subgroup of a symmetric group. The symmetric groups themselves are good examples.

Carsten S
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  • maybe a silly question, but is what you say in fact restricted to finite groups? – David Holden Dec 11 '13 at 08:54
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    @DavidHolden, that is Cayley's Theorem and I don't think it is restricted to finite groups. – David Dec 11 '13 at 08:56
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    Indeed it's not restricted to finite groups: every group $G$ acts by (e.g.) right multiplication on itself, giving an embedding of $G$ in the symmetric group on $G$. – Shane O Rourke Dec 11 '13 at 08:59
  • @Shane (& David - the software won't let me put more than one addressee on a comment) thank you. does that mean the alternating group has index 2 even if the group is infinite? – David Holden Dec 11 '13 at 09:05
  • David has given the name of the theorem and Shane its proof :) Maybe I should have stated it as „every finite group is isomorphic to a subgroup of a finite symmetric group“. The infinite case is also true, but I somehow find it less interesting. – Carsten S Dec 11 '13 at 09:12
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    @DavidHolden How do you define even and odd for permutations of an infinite set? – Carsten S Dec 11 '13 at 09:14
  • @Carsten good point, thanks. i had an (erroneous) picture of a group which contained only the permutations of an (arbitrarily large) finite subset. i suppose, then, that what i had in my mind does exist as a subgroup of the full symmetric group of permutations of an infinite set, and that only this subgroup (say $S^$) has an alternating subgroup ($A^$?) of index 2. the finiteness condition suggests $S^*$ must be a normal subgroup of $S$. if so, i have now a difficulty picturing the factor group. is there some conceptual analogy with the construction of function germs? – David Holden Dec 11 '13 at 10:29
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    @DavidHolden: Scott's Group Theory textbook describes these groups very well, including their normal structure. – Jack Schmidt Dec 12 '13 at 14:49
  • Most of the terminologies used in the answers above I do not understand clear enough. Can some one help giving an example of a noncyclic finite group by citing the set and its operation. Modulo prime multiplication group forms a cyclic group with multiple generating elements. modulo addition any natural number gives another example of finite cyclic groups.Is there such a simple example of finite group for noncyclicity? – Seetha Rama Raju Sanapala Oct 03 '16 at 04:23