I know that many kind of groups can be represented by matrices; for example: rotation groups can be represented by matrices. Especially all elements of rotation groups can be represented by orthogonal matrices with determinant positive one. Also Permutation group can be represented by matrices.

But I need to know:

  • are there any kind of group which can not be represented by matrices?

  • Or is there any kind of group which does not have a representation?

Can you show me sample?

  • $\begingroup$ http://en.wikipedia.org/wiki/Regular_representation $\endgroup$ – blue Mar 10 '14 at 8:29
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    $\begingroup$ As in your other question, I think a few of us is very confused at what you mean by "representing a group", and I think this leads to a mess in the answers and comments below. In representation theory, a representation of a group is a homomorphism $\rho:G\to GL(V)$ for some vector space $V$, can you confirm this is what you mean? $\endgroup$ – Aaron Mar 10 '14 at 13:26

There are infinite groups which cannot be represented by finite-dimensional matrices over any commutative ring. If a group can be represented by matrices in such a way then it is called linear$^{\ast}$. Not all groups are linear.

A group $G$ is residually finite if for every element $g\in G$ there exists a homomorphism $\phi_g: G\rightarrow F_g$ onto a finite group $F_g$ such that $\phi_g(g)$ is non-trivial. Equivalently, for every element $g\in G$ there exists an action of $G$ on a finite object such that the action of $g$ is non-trivial. It is a rather famous result of Malc'ev that finitely generated linear groups are residually finite. This allows you to conjure up non-linear groups almost at will!

Non-Hopfian groups One way of constructing finitely-generated, non-residually finite groups is to construct finitely generated groups which have a surjective endomorphism $G\rightarrow G$ which is not an isomorphism. Such groups are called non-Hopfian, and it is a result of (again) Malc'ev that these groups are non-residually finite. See this Math.SE answer of mine for the proof, and this one for examples of such groups (the main example is the group $\langle a, b; b^{-1}a^2b=a^3\rangle$).

Simple groups A second way of constructing finitely-generated, non-residually finite groups is to construct finitely generated infinite simple groups. Examples of such groups are Thompson's group's $T$ and $V$ (these can be realised as groups acting on the unit interval in very natural ways - see these notes or this answer of mine) and Tarski monster groups.

Higman's group The group $G=\langle a, b, c, d; a^{-1}ba=b^2, b^{-1}cb=c^2, c^{-1}dc=d^2, d^{-1}ad=a^2\rangle$ was the first example of a finitely generated, infinite group with no finite quotients. This clearly implies that $G$ is not residually finite. Higman's paper is a joy to read$^{\dagger}$, and in it he points out that $G$ can easilly be used to construct finitely generated, infinite simple groups (his paper was pre-Thompson's groups, and pre-Tarski monsters) - taking a maximal normal subgroup $N$ of $G$, $G/N$ must be simple!

$^{\ast}$Perhaps this definition really requires field not commutative ring, but everything in this answer works for the more general commutative ring definition.

$^{\dagger}$Higman, Graham (1951), A finitely generated infinite simple group, J.Lon. Math. Soc.

  • $\begingroup$ I agree that the definition of linear is more natural over a commutative ring than a field. Anyway I don't know an example of finitely generated group that would be linear over a commutative ring, but not over a reduced commutative ring (or equivalently over a finite product of fields). $\endgroup$ – YCor Sep 18 '14 at 11:36
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    $\begingroup$ Btw, I would say that today, the interest of Higman's group is not that it has infinite simple quotients (because as you mention there are many others, more natural than a quotient obtained by finding a non-explicit maximal normal subgroup), but that it's finitely presented with an explicit presentation and has no trivial quotient, and that in addition it is very rich in terms of quotients (it's SQ-universal). $\endgroup$ – YCor Sep 18 '14 at 11:40

A representation of a group $G$ is a morphism $\rho:G\rightarrow\text{GL}(V)$ where $V$ is a vector space over a field $K$.

There are always at least the following two representations:

  • the trivial representation, where $\rho(g)=\text{id}_V$ for all $g\in G$.

  • the regular representation, where $V$ is the space of $K$-valued functions on $G$ and $\rho(g)\phi(x)=\phi(xg)$ for all $\phi\in V$ and for all $x, g\in G$.

EDIT: Also note that whenever $H$ is normal in $G$ any representation of $G/H$ can be lifted to a representation of $G$ by composing with the quotient homomorphism $G\rightarrow G/H$.

In particular if the quotient $G/H$ is abelian this gives rise to lots of representations of $G$ since abelian groups have many representations, starting with characters (i.e. $1$-dimensional representations)

  • $\begingroup$ So you mean that all kind of groups have a representation? even minimum two representation! right? $\endgroup$ – IremadzeArchil19910311 Mar 10 '14 at 8:47
  • $\begingroup$ Andrea, is this still valid if $G$ is infinite? $\endgroup$ – Nicky Hekster Mar 10 '14 at 8:48
  • $\begingroup$ Of course this is valid also when $G$ is infinite. Mind, though, that in that case $\dim_K(V)$ is also infinite. $\endgroup$ – Andrea Mori Mar 10 '14 at 9:06
  • $\begingroup$ @Andrea see user1729's answer, this is what I meant ... $\endgroup$ – Nicky Hekster Mar 10 '14 at 10:07
  • $\begingroup$ @Nicky The difference is that I am assuming the matrices are finite-dimensional but Andrea Mori is not. I have modified the first paragraph of my post to make this clear. $\endgroup$ – user1729 Mar 10 '14 at 10:33

As you stated, every (finite) permutation group can be represented by matrices (take $G=S_n$ and represent the permutations by permutation matrices (the permutations act naturally on a vector space of dimension $n$ over some field $K$)). Subsequently, every finite group can be embedded in a permutation group (a Theorem of Cayley), since it acts on itself by left (or right) multiplication.

  • $\begingroup$ I need to know which group can not be represented by representation? $\endgroup$ – IremadzeArchil19910311 Mar 10 '14 at 8:35
  • $\begingroup$ It follows that each finite group has a reresentation. I concluded from your question that you were talking about finite groups. $\endgroup$ – Nicky Hekster Mar 10 '14 at 8:42
  • $\begingroup$ what about infinite groups? So you mean that if there is a group which can not be represented by representation, then that group might be infinite groups?! $\endgroup$ – IremadzeArchil19910311 Mar 10 '14 at 8:43

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