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.
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$\begingroup$ 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" $\endgroup$– David HoldenDec 11, 2013 at 8:36
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$\begingroup$ Dihedral groups, symmetric groups, alternating groups, groups of matrices over finite fields. $\endgroup$– Gerry MyersonDec 11, 2013 at 8:37
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$\begingroup$ @David, that's a direct product. $\endgroup$– Gerry MyersonDec 11, 2013 at 8:37
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1$\begingroup$ 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. $\endgroup$– Gerry MyersonDec 11, 2013 at 9:07
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1$\begingroup$ 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? $\endgroup$– DavidDec 11, 2013 at 9:10
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2 Answers
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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.
answered Dec 12, 2013 at 9:39
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$\begingroup$ link to wikipedia page? en.wikipedia.org/wiki/Klein_four-group as their explanation (and the Cayley table really helps!) $\endgroup$– Jason SJun 22, 2017 at 14:59
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1$\begingroup$ 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.) $\endgroup$– user1729Jan 25, 2019 at 11:41
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$\begingroup$ The Klein $V$ group is called such because the German term for the number 4 is "vier", $\endgroup$ Sep 25, 2019 at 4:22
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Every finite group is isomorphic to a subgroup of a symmetric group. The symmetric groups themselves are good examples.
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$\begingroup$ maybe a silly question, but is what you say in fact restricted to finite groups? $\endgroup$ Dec 11, 2013 at 8:54
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1$\begingroup$ @DavidHolden, that is Cayley's Theorem and I don't think it is restricted to finite groups. $\endgroup$– DavidDec 11, 2013 at 8:56
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1$\begingroup$ 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 $G$. $\endgroup$ Dec 11, 2013 at 8:59
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3$\begingroup$ @DavidHolden How do you define even and odd for permutations of an infinite set? $\endgroup$ Dec 11, 2013 at 9:14
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1$\begingroup$ @DavidHolden: Scott's Group Theory textbook describes these groups very well, including their normal structure. $\endgroup$ Dec 12, 2013 at 14:49