77
$\begingroup$

Can anybody provide some examples of finite nonabelian groups which are not symmetric groups or dihedral groups?

$\endgroup$
12
  • 7
    $\begingroup$ How about matrix groups over finite fields? $\endgroup$
    – muzzlator
    Mar 12, 2013 at 6:00
  • 13
    $\begingroup$ Quaternions $Q_{8}$. $\endgroup$ Mar 12, 2013 at 6:01
  • 15
    $\begingroup$ Every finite group is isomorphic to a subgroup of a symmetric group. So, no really new examples are possible. $\endgroup$
    – Dan Shved
    Mar 12, 2013 at 6:01
  • 6
    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Nikolaj-K
    Mar 12, 2013 at 8:36
  • 5
    $\begingroup$ community wiki? $\endgroup$
    – user58512
    Mar 12, 2013 at 12:32

10 Answers 10

162
+100
$\begingroup$

Here's a bunch of examples I've collected from my notes, group theory texts, and various places around the Internet. This has become somewhat of a treatise, but nonetheless, I hope you and others enjoy them.

Familiar generalizations.

  • Generalized dihedral groups, denoted $\mathcal{D}(A)$ or $\operatorname{Dih}(A)$, are formed by letting an involution (element of order $2$) act on an arbitrary abelian group $A$ by inversion. So, $D_{2n}=\operatorname{Dih}(\mathbb{Z}/n\mathbb{Z})$.

  • Another sister of the dihedral groups are the semidihedral groups. These are all $2$-groups which behave just a tiny bit different from their dihedral counterparts (see the presentation).

  • The quaternion group $Q_8$ (not to be confused with the related-but-not-identical quaternion algebra) is a Hamiltonian group of order $8$. It's made of three normal subgroups $\langle i \rangle$, $\langle j \rangle$, and $\langle k \rangle$, which intersect at the order $2$ center $Z(Q_8)=\langle -1 \rangle$. $Q_8$ is the first counterexample you should try whenever you think anything might be true. (one example)

  • The generalized quaternion groups, denoted $Q_{4n}$ for $n\geq 2$, naturally extend the group presentation of $Q_8$ to higher orders. The important thing about generalized quaternion groups is that they have a unique element of order $2$, whereas all other $p$-groups with unique elements of order $p$ are cyclic.

    • These can be most easily understood as the quotient of the semidirect product $\mathbb{Z}_{2^n}\rtimes \mathbb{Z}_4$ by the subgroup $\langle(2^{n-1},2)\rangle$.

    • You can also realize $Q_{4n}$ as the subgroup of $\mathbb{H}^\times$ generated by $\cos(\pi/n)+\mathbf{i}\sin(\pi/n)$ and $\mathbf{j}$, where $\mathbb{H}$ denotes the (division) ring of Hamiltonian quaternions (with generators $\mathbf{i},\mathbf{j},$ and $\mathbf{k}$, under the usual relations). I found this construction very interesting

I'll stop here for a moment to note that the dihedral groups, semidihedral groups, and generalized quaternion groups constitute all $2$-groups of maximal class, which makes them super important to know for people who study $p$-groups.

  • We can even generalize the generalized quaternions. Replacing $\mathbb{Z}_{2^n}$ in the above definition with an arbitrary cyclic group $A$ of even order $n$ gives us the dicyclic groups, denoted $\operatorname{Dic}(A)$ or $\operatorname{Dic}_n$. We can get even further into outer space and define the generalized dicyclic groups $\operatorname{Dic}(A,y)$ by allowing $A$ to be any abelian group with even order. (For generalized dicyclic groups one must specify which element $y\in A$ of order $2$ behaves like the unique element of order $2$ in the generalized quaternion group. You can read about that here.)

Matrix groups.

Scary groups.

  • The sporadic groups are the scariest things which lurk outside of the happy little bubble that is solvable groups. This is where the Monster lives. In most cases, understanding the structure of these groups requires some esoteric machinery, so I would avoid these until you're really good at group theory.

By the way, any nonabelian simple group is an alternating group, a $\operatorname{PSL}$ or other group of Lie type, or a sporadic group. Figuring that out took about $110$ years of focused sorcery, which is fun to read about if you like math history.

Frobenius groups.

  • Frobenius groups are made of two parts: a normal subgroup $N$, called the Frobenius kernel, which is acted upon by a subgroup $K$, the Frobenius complement, with the defining property that $K$ acts on $N$ fixed point freely - that is, $n^k\ne n$ for all non-identity elements $n\in N,k\in K$. One example is $S_3$, for which the kernel is generated by the $3$-cycles and the complement is generated by (your choice of) a $2$-cycle. The structure of Frobenius groups is well understood - as it turns out, $N$ must be nilpotent, and the Sylow subgroups of $K$ must be either cyclic or generalized quaternion.

Frobenius groups are also the subject of a major theorem in representation theory which proves an alternative definition is equivalent to the standard one. I personally don't find this alternative definition particularly enlightening, but the proof of this theorem demonstrates how powerful representation theory can be.

  • $2$-Frobenius groups are groups $G$ which have some subgroup $F$ which is a Frobenius group with kernel $K$ which is normal in $G$ for which $G/K$ is also a Frobenius group. An example of this is $S_4$, where $F=A_4$ (the subgroup $K$ generated by the permutions $(12)(34)$ and $(13)(24)$ are acted upon fixed point freely by $\langle (123) \rangle$) and $G/K\cong S_3$.

Frobenius and $2$-Frobenius groups are important for element orders. Whenever a solvable group $G$ lacks an element of order $pq$, where $p$ and $q$ are primes dividing $|G|$, the Hall $\{p,q\}$ subgroups of $G$ (subgroups of order $p^aq^b$ with index coprime to their order) are either Frobenius or $2$-Frobenius.

$2$-Frobenius groups were also important in the proof of the Odd Order theorem, which derived a contradiction by showing that a non-solvable group with odd order would have to contain certain $2$-Frobenius groups in a way that was not possible.

  • Zassenhaus groups are somewhat related to Frobenius groups in that they are defined by counting fixed points of their action on a set. (I'll include more info on these soon.)

Cool families of $p$-groups.

$p$-groups are groups of prime power order $p^n$. We already saw some families of $2$-groups in the first section; here, we will discuss families of $p$-groups of where $p$ is not necessarily even.

  • A special group is a $p$-group $G$ in which the center, derived, and Frattini subgroups coincide ($Z(G)=G^\prime=\Phi(G)$). If these subgroups are cyclic of order $p$, we call the group extraspecial. The combination of these properties gives rise to some very interesting representation-theoretic behavior. In special groups, the commutator map from $G/Z(G)\times G/Z(G)\rightarrow G^\prime$ satisfies $[g,hk]=[g,h][g,k]$, so in extraspecial groups this may be considered as bilinear form over $\mathbb{F}_p$ (which, as it turns out, is skew-symmetric, alternating, and non-singular). If $p$ happens to be $2$, we can additionally show that the squaring map is a quadratic form. You can read more about this in $\oint 3.10$ of Wilson's Finite Simple Groups, which can be seen in a preview on the web. Note also that it is particularly easy to classify the conjugacy classes of extraspecial groups, so they often make good testing grounds for conjectures in representation theory.

Some favorites from my toolbox.

  • $M_{16}$, the group with the coolest name, is the smallest example of a group with two distinct isomorphic characteristic subgroups, and thus another great counterexample group. $M_{16}$ is pretty close to being abelian - it is just $\mathbb{Z}_2\times \mathbb{Z}_8$ with a little "twist."

  • The octahedral group is the symmetries of an octahedron. I'm partial to the binary octahedral group $2\mathcal{O}$, which has order $48$ and is constructed by replacing certain elements of $\operatorname{GL}_2(\mathbb{F}_3)$ with scalar multiples in $\operatorname{GL}_2(\mathbb{F}_9)$. (I recently heard from a coauthor that Marty Isaacs refers to this group "fake $\operatorname{GL}(2,3)$", which I think is hilarious.)

  • $\Gamma$ groups are the affine transformation version of $\Gamma\operatorname{L}_n(\mathbb{F}_q)$. I like taking small subgroups of one dimensional $\Gamma$ groups when I want a very nonabelian group in which I can still easily write out arithmetic.

  • My second favorite matrix group is the Valentiner group, isomorphic to $\operatorname{PGL}_3(\mathbb{F}_4)$. My favorite sporadic group is the Thompson group.

  • I'm a big fan of making Frobenius groups of the form $\mathbb{Z}_{p}\rtimes \mathbb{Z}_q$ (and $2$-Frobenius groups of the form $(\mathbb{Z}_{p}\rtimes \mathbb{Z}_q)\rtimes \mathbb{Z}_r$), where $p,q,r$ are primes (necessarily) satisfying $p\equiv 1 \pmod q$ (and $p\equiv 1 \pmod {qr})$. Need an easy nonabelian group of order $21$? Boom, done.

  • Another particularly nice, easy class of Frobenius groups are those of the form $C_p\rtimes C_{p-1}$, where $p$ is an odd prime, also known as the holomorph of $C_p$, $\operatorname{Hol}(C_p)$. These groups arise naturally in Galois theory as the Galois group of $\mathbb{Q}\left(2^{1/p},\xi\right)$, the splitting field of $x^p-2$ over $\mathbb{Q}$. (Here, $\xi$ is a $p^\text{th}$ root of unity.) In this case, the most obvious set of generators is $$\sigma:\left\{\begin{array}{l}2^{1/p}\mapsto \xi 2^{1/p}\\ \xi\mapsto \xi\end{array}\right.\hspace{15pt}\text{and}\hspace{15pt}\tau:\left\{\begin{array}{l}2^{1/p}\mapsto 2^{1/p}\\ \xi \mapsto \xi^m\end{array}\right.$$ where $m$ has order $p-1$ mod $p$.

  • For an abstraction of the previous example, check out Z-groups, groups with all Sylow subgroups are all cyclic. These are related to extraspecial groups and Zassenhaus groups, and tend to be good examples in character theory. Like most things with well-understood character theory, Z-groups came up a lot in the papers leading up to the Classification theorem.

  • Double and triple covers of alternating groups show up as counterexamples in finite group theory often.

Make your own.

  • You can take wreath products of any of the above nonabelian groups to get very strange looking groups indeed. (Just the other day $(S_p\wr S_p)\wr S_p$ came up in chat.)

  • If you know something about the automorphisms of a group $G$, you can take semidirect products $G\rtimes H$ with other groups $H$ (usually those that embed into $\operatorname{Aut}(G)$). In particular, holomorphs can be an easy way to make new groups with easy arithmetic.

  • If two groups $H$ and $K$ have isomorphic central subgroups $Z_1\leq Z(H)$ and $Z_2\leq Z(K)$, you can take the (external) central product of $H$ and $K$, written $H{\small\text{ Y }} K$. This is done by identifying points in $Z_1$ with the corresponding points in $Z_2$, then taking the quotient of $H\times K$ by this diagonal central subgroup. (Explicitly, if $\theta:Z_1\rightarrow Z_2$ is an isomorphism, let $Z=\{(z,\theta(z)^{-1}):z\in Z_1\}$, and $H{\small\text{ Y }} K=\left(H\times K\right)/Z$.) There's a rich theory behind central products, and you can make some nifty things happen with them. (example)

  • If you want to give a group theorist a headache, take a direct product with one component from each of the above categories, wreath it with itself, and ask about the structure of its Sylow $2$-subgroup.

$\endgroup$
10
  • 3
    $\begingroup$ A great list! What's the smallest non-abelian group that is not on it (if you throw in the cyclic groups)? $\endgroup$
    – yatima2975
    Mar 13, 2013 at 21:19
  • 1
    $\begingroup$ @yatima2975 Due to the inclusion of semidirect/wreath products that's a pretty hard question. I'll think about it and get back to you. $\endgroup$
    – Alexander Gruber
    Mar 14, 2013 at 8:24
  • $\begingroup$ Are central products covered? If not, then there's a group of order 16 that's missing. I'm looking for a toolbox to build up groups from smaller ones, without talking (too much) about subgroups and quotients. Keep me updated! $\endgroup$
    – yatima2975
    Mar 14, 2013 at 14:22
  • 1
    $\begingroup$ @Steve It's got a cool uniqueness proof! Also it isn't talked about too much, and I have a thing for underdogs. $\endgroup$
    – Alexander Gruber
    Apr 3, 2013 at 5:03
  • 1
    $\begingroup$ @yatima2975 Actually one can make that group as $Q_8\rtimes \mathbb{Z}_2$ formed via the automorphism $y\mapsto yz$, using the presentation $Q_8=\langle x,y,z|x^2=z,y^2=z,y^x=yz\rangle$. $\endgroup$
    – Alexander Gruber
    Apr 13, 2013 at 0:43
12
$\begingroup$

The 168 element symmetry group of the Fano plane, which is the smallest example of a simple group that is neither abelian (cyclic) nor covered by meh's answer (alternating). And it is of course onlly the smallest member of a bigger family ;)

$\endgroup$
10
$\begingroup$

The symmetry group of the rubiks cube is non-abelian, otherwise it would be easy to solve one!

$\endgroup$
2
  • 3
    $\begingroup$ Twist right side then twist top $\neq$ twist top then twist right side (where a twist is a quarter turn rotation in a given direction you set beforehand). $\endgroup$
    – user1729
    Mar 12, 2013 at 12:46
  • 5
    $\begingroup$ Also, as a corollary you get that there is no single sequence $T$ of twists such that a repeated application of $T$ will solve your cube from any position. The existence of such a sequence $T$ would mean that the group is cyclic, and so abelian. I still find this amazing (as a cute application of group theory to the real world)! Also, if you want to impress your friends, pick up a cube and do a fixed sequence of twists. Repeat. Repeat. Repeat... You will end up back at the beginning. You will become more attractive to the opposite sex as a result. $\endgroup$
    – user1729
    Mar 12, 2013 at 13:06
7
$\begingroup$

Take any group $G$ and any symmetric group $S_n$ with $n\ge 3$. Then $G \times S_n$ is not abelian, even if $G$ is abelian.

$\endgroup$
5
$\begingroup$

The quaternion group is non-abelian and of order 8.

$\endgroup$
3
$\begingroup$

Feels like I'm cheating here, but Alternating Groups, which are derived from symmetric groups.

$\endgroup$
3
$\begingroup$

You can also construnt non-abelian finite groups from finite abelian groups. For example, any cyclic group $C_{n}$ will be abelian. However, the wreath product $C_{n}\wr C_{n}$ is an example of a non-abelian finite group (well strictly speaking I should say for $n\geq 2$ here).

$\endgroup$
2
$\begingroup$

you can define a group by define the product of the elements and define some two elements not to commute with each other , but the order of this group must be bigger than or equal to 6 because any group of order less than 6 is ableian , but important thing which you should know is that cayley's theorem says that any finite group of order $n$ is isomorphic to subgroup of $S_n$ , so if you look for non abelian group which is not isomorphic to symmetric group or permutation group then , look for in infinte groups , for example , let the set $G$ = $R$ " the reals " , define binary operation on $G$ as follows ,$ a*b = a^{b}$ for all $a,b$ $\in$ $G$ , you can check that this make a group which is not abelian , there is an idenity and inverses and it's assiocative and it's closed . so this group is non-abelian

$\endgroup$
2
  • 1
    $\begingroup$ Note that the question is on finite groups. $\endgroup$
    – azimut
    Mar 12, 2013 at 18:35
  • $\begingroup$ @azimut , oh yeah , i didn't watch that :) .. i typed what i think is true about the problem in general , if i made some mistake , plz correct it :) $\endgroup$ Mar 12, 2013 at 22:15
1
$\begingroup$

A subfamily of triangle groups.

$\endgroup$
0
1
$\begingroup$

I will employ here again the same web resource by Eric Moorhouse I linked yesterday in this answer to a related question.

This finite group and this other have both order $27$: since there's no $n$ such that $n!=27$, they are not symmetric; since there's no $m$ such that $2m=27$, they are not dihedral. Just a glance at the bottom right corner of the two Cayley tables to see they are both non commutative.

The two example given are not isomorphic to each other and have different exponent.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .