# Examples of infinite nonabelian groups not GL_n(G)?

1. Sep 9, 2015

### jackmell

Hi,

I was trying to identify some infinite non-abelian groups other than $GL_n(G)$ and also other than contrived groups such as the group:

$G=\big<r,s : r^2=s^3=1\big>$

as per:

https://www.physicsforums.com/threa...and-nonabelian-given-its-presentation.643968/

and I was wondering if someone here could give me some examples that I could understand. Does not seem to be common and I didn't find any with a cursory search but may have to just search more.

Thanks,

Jack

Last edited: Sep 9, 2015
2. Sep 9, 2015

### WWGD

Have you considered Quaternions?

3. Sep 9, 2015

### jackmell

But that's only a finite group $Q=\{\pm 1, \pm i, \pm j, \pm k\}$ although perhaps I'm not fully understanding what can be done with this group.

Edit:

Maybe though I can consider the group $G=\{a+bi+cj+dk\; :\; (i,j,k)\in Q, (a,b,c,d)=\mathbb{R}\}$. But then wouldn't multiplication still be commutative? Looks to be the case.

Not so. I can let $h_1=i$ and $h_2=j$ then $h_1 h_2\neq h_2 h_1$. I'm getting it.

But now I have to figure out what the inverse of $g\in G$ is. I assume that's the conjugate $(a-bi-cj-dk)$? Need to check that. Just haven't done much with quaternions.

Last edited: Sep 9, 2015
4. Sep 9, 2015

### WWGD

Right, the first is just a finite generating set and not the group itself. And you should consider the relations, e.g.. $i^2=-1$. Maybe viewing them as rotations will help you get a more intuitive view, a better grasp of them --please let me know when you do, because I dont have it myself ;).

5. Sep 10, 2015

### lpetrich

The group of quaternions is equivalent to the group U(2), the group of 2*2 unitary matrices. For unit quaternions, it's SU(2), the U(2) matrices with determinant 1.

More generally, however, GL(n,C) > U(n) > SU(n)

But by mapping complex numbers a+b*i onto a*{{1,0},{0,1}} + b*{{0,-1},{1,0}} = {{a,-b},{b,a}}, one finds that GL(n,C) is a subgroup of GL(2n,R).

6. Sep 10, 2015

### lpetrich

More generally, the OP seems to be asking if there is some infinite group that does not map onto some subgroup of GL(n,C) for some n. Such a subgroup is called a "representation" of the group, and the OP is asking if there is some group without representations. Strictly speaking, every group has a representation: the identity one, the identity matrix for every element of the group. But that case aside, the trivial case, the interesting case is a "faithful" representation, where distinct group elements map onto distinct representation elements.

Thus, the OP's question is whether there is some infinite group that has no faithful representations, or even no nontrivial representations.

In the finite case, that is easy to prove, by constructing the "regular representation", D(a)x,y = 1 if x = a*y, 0 otherwise. It is not difficult to show from that definition that D(a).D(b) = D(a*b).

I don't know if anyone has proved that for infinite groups, however.

7. Sep 10, 2015

### WWGD

I assumed when you mention a group, it is up to isomorphism. Representations are usually just homomorphisms.

8. Sep 10, 2015

### Ben Niehoff

9. Sep 10, 2015

### pasmith

Consider the set of bijections from [0,1] to itself under the operation of composition of functions. This group is infinite and not abelian.

10. Sep 10, 2015

### WWGD

Following up on pasmith's example, consider the mapping class group of a topological space, also under composition. These are automorphisms (self -homeomorphisms), often diffeomorphims, up to isotopy.https://en.wikipedia.org/wiki/Mapping_class_group . Many of these are infinite, but not all.

Last edited: Sep 10, 2015
11. Sep 11, 2015

### jackmell

Hi,

Isn't that just isomorphic to a set of permutations? I'm not sure but if I define:

$\phi: \{ \sigma: [0,1]\to [0,1] \text{ is a bijection}\}$

then I could write $\phi\cong S_{\infty}$ and I know that the symmetric group is non-abelian.

Also the reason I asked was related to a homework problem which I didn't want to ask you guys for help since wabbit had already helped me with the first part:

but since I've already turned it in maybe you guys could tell me what I did wrong because I don't think I got it right:

Give an example of a group in which the normalizer, $N_G(X)$ is not equal to the set $\{g\in G : gxg^{-1}\in X;\forall\; x\in X\}$ for some set $X\subseteq G$.

My professor had briefly suggested in class using $GL_n(F)$ but I wanted to try something different so I tried that quaternion group above and let $X=\{x_0, x_1, x_2,\cdots,x_n,\cdots\}$ with $x_0$ some non-identity element in G and then $x_1=g_0xg_0^{-1}, x_2=g_0x_1g_0^{-1}, \cdots$, again for some non-identity $g_0$ element in G. Then I argued that almost certainly we have the following:

$N_G(X)=\{1\}$
but
$\{g\in G : gxg^{-1}\in X\;\forall\; x\in X\}=\{1,g_0\}$

But I couldn't prove it so left it at "almost certainly". Maybe $\{ \sigma: [0,1]\to [0,1] \text{ is a bijection}\}$ would have been a better choice.

Last edited: Sep 11, 2015
12. Sep 18, 2015

### lpetrich

Yes, it's a faithful representation that's an isomorphism, one whose kernel contains only the identity element.

13. Sep 18, 2015

### lpetrich

So that's a counterexample to what one would expect from a finite group. The regular representation of a finite group can easily be shown to be a faithful matrix representation, but that concept does not carry over very well into infinite groups. For countably infinite groups, one might define an infinite-dimensional matrix as an extension of finite-dimensional ones, but I don't think that one can do that for uncountably infinite groups. One might generalize a matrix into function composition, however, and get a function-composition generalized representation for an uncountably infinite group. Matrix multiplication -> function composition:

A.B.x -> A(B(x))

An invertible matrix would generalize to a bijection, an invertible function. Function composition is associative, and there is an identity function. So function composition in general is a monoid. With only invertible functions, it is a group.