# Can every group be represented by a group of matrices?

Can every group be represented by a group of matrices?

Or are there any counterexamples? Is it possible to prove this from the group axioms?

-
What are our limits? Any singleton is trivially a 1 by 1 matrix, and if we define the matrix operation to align with the group operation then it's true. Or, we could deal with diagonal matrices, all of whose elements are the same. Then the matrix operation would do repeated group operations (one per element), and again this gives something isomorphic to the group. –  mixedmath Nov 24 '11 at 18:00
For any field $F$ if you take a group of larger cardinality then there is no way the group could be represented by finite matrices over $F$. –  Carl Mummert Nov 25 '11 at 11:26
If you're willing to allow linear transformations on infinite-dimensional vector spaces, then every group can be realized as a set of linear transformations. –  arsmath Feb 12 at 12:27

## 5 Answers

Every finite group is isomorphic to a matrix group. This is a consequence of Cayley's theorem: every group is isomorphic to a subgroup of its symmetry group. Since the symmetric group $S_n$ has a natural faithful permutation representation as the group of $n\times n$ 0-1 matrices with exactly one 1 in each row and column, it follows that every finite group is a matrix group.

However, there are infinite groups which are not matrix groups, for example, the symmetric group on an infinite set or the metaplectic group.

Note that every group can be represented non-faithfully by a group of matrices: just take the trivial representation. My answer above is for the question of whether every group has a faithful matrix representation.

-
I believe that the metaplectic group has no faithful continuous finite-dimensional representations, but is it true that it has no faithful finite-dimensional representations as an abstract group? –  Qiaochu Yuan Nov 24 '11 at 18:20

It is not true that every group can be represented by a group of finite-dimensional matrices (say over $\mathbb{C}$). The groups that can are called linear. There are many examples of non-linear groups; here is a relatively simple one.

Claim: The group $(\mathbb{Z}/2\mathbb{Z})^{\infty}$ is not linear.

Proof. Suppose to the contrary that there exists a faithful representation $(\mathbb{Z}/2\mathbb{Z})^{\infty} \to \text{GL}_n(\mathbb{C})$ for some $n$. In particular, for arbitrarily large $m$, there exists a faithful representation $(\mathbb{Z}/2\mathbb{Z})^m \to \text{GL}_n(\mathbb{C})$. We can conjugate this to a representation into $U(n)$ and then simultaneously diagonalize to obtain a representation into $\mathbb{T}^n$. But the subgroup of elements of $\mathbb{T}^n$ of order $2$ is $(\mathbb{Z}/2\mathbb{Z})^n$, so the representation cannot be faithful if $m > n$; contradiction.

-
Just curious, can you elaborate on the second last sentence a little more? –  Eric Naslund Nov 24 '11 at 21:05
@Eric: every complex representation of a finite abelian group decomposes into a direct sum of $1$-dimensional representations, so all of the elements are simultaneously diagonalizable. –  Qiaochu Yuan Nov 25 '11 at 1:37
I know that, but what does $\mathbb{T}^n$ mean here? –  Eric Naslund Nov 25 '11 at 7:38
@Eric: I believe it means the $n$-dimensional torus. –  Zev Chonoles Nov 26 '11 at 18:03

For finite groups, the answer is yes: every finite group is isomorphic to a subgroup of $Gl(n)$ for some $n$ large enough. In fact, we can do slightly better and embed it into $SO(n)$ for $n$ large enough.

The idea: Every group acts on itself, by, say, left translations. This gives an embedding $G\rightarrow S_{|G|}$ into the symmetric group on $|G|$ letters. Hence, every group is a subgroup of some $S_k$ for $k$ large enough.

Thus, we just need to show that we can embed every $S_n$ into $GL_n$ for $n$ large enough. But $S_n$ acts on $\mathbb{R}^{n}$ by permuting the coordinates. It's easy to see that this action is linear, hence defines an embedding $S_n\rightarrow GL(n)$.

If you insist on positive determinant, one can include $GL(n)$ into $Gl(n+1)$ sending a matrix $A$ to $\operatorname{diag}(A, \operatorname{sign}(\det(A)))$.

Finally, to embed orthogonally, pick your favorite inner product $\langle, \rangle$ on $\mathbb{R}^n$ and define $\langle x, y\rangle_1 = \sum_{g\in G} \langle g(x),g(y)\rangle$. It's easy to see that this new metric is $G$ invariant, so $G$ acts by isometries in this metric, hence, $G$ embeds into $SO(n)$ for $n$ large enough.

The story for infinite groups is more complicated. Of course, there are groups of any preassigned cardinality, while any subset of $Gl(n)$ can have cardinality at most that of the real numbers. So, no sufficiently large group can embed into $Gl(n)$.

If you mix a bit of topology, then, for example, every compact Lie group embeds into $SO(n)$ for $n$ large enough, but the proof is not trivial. Noncompact Lie groups need not embed into $GL_(n)$ for any $n$. For example, I believe that the universal cover of $Sl(2)$ is such an example.

-
It is not actually obvious that there are groups of any preassigned cardinality (what is clear is that there are groups of at least any preassigned cardinality). The statement that every set admits a group structure is equivalent to the axiom of choice (mathoverflow.net/questions/12973/…). –  Qiaochu Yuan Nov 24 '11 at 18:18
@Qiaochu: I was thinking of taking the direct sum of sufficiently many $\mathbb{Z}/2\mathbb{Z}$, not that every set has a group structure. –  Jason DeVito Nov 24 '11 at 18:21
The regular representation is probably the easiest to see. We can associate to a finite group of order $n$, $G=\{g_1,g_2,\dots,g_{n}\}$ a vector space $V_G$ with basis $\{e_{g_1},e_{g_2},\dots,e_{g_{n}}\}$. Now define linear maps $T_g\in \text{L}(V_G)\subset\text{GL}_{n}(\mathbb{C})$ by $T_ge_h=e_{gh}$. The map $\varphi:G\rightarrow \text{GL}_n(\mathbb{C})$, $\varphi(g)=T_g$ is an isomorphism of groups. –  Jp McCarthy Nov 24 '11 at 18:50
@Jason: is it obvious that you get every cardinality that way? –  Qiaochu Yuan Nov 24 '11 at 19:04
@Qiaochu: I didn't think through it in great detail. But, I know that (at least assuming choice) for each cardinal $\kappa$, $\oplus_{\kappa} \mathbb{Z}/2\mathbb{Z}$ has cardinality $\kappa$ (since it's equivalent to counting the finite subsets of $\kappa$). Does counting finite subsets of a cardinal number require choice? Probably, but I'm not sure how much. Either way, I find it more intuitive that "every set has a group structure", which, as you link to above, is completely equivalent to AC. –  Jason DeVito Nov 24 '11 at 19:15

Your question is slightly vague, but...this paper discusses groups which are counter-linear: a group $G$ is counter-linear if for every field $K$ and every positive integer $n$ the only homomorphism $G \rightarrow \operatorname{GL}_n(K)$ is the trivial homomorphism. In particular, counter-linear groups exist! examples of counter-linear groups are given. (See also Jack Schmidt's comment below.)

In group theory one often says that a group is linear over a field $K$ if it admits a faithful $n$-dimensional $K$-representation, i.e., an injective homomorphism $\rho: G \hookrightarrow \operatorname{GL}_n(K)$, for some positive integer $n$. In particular counter-linear groups are not linear over any field, but it is easier to give examples of the latter. For instance, Qiaochu's group $(\mathbb{Z}/2\mathbb{Z})^{\infty}$ is not linear over any field. I feel like I gave a similar answer on MO at one point, but I just searched unsuccessfully for it. Oh, well. Anyway, an argument for the general case can be obtained by counting conjugacy classes of involutions in $\operatorname{GL}_n(K)$: this is a good linear algebra exercise.

Note that if you allow "infinite dimensional matrices" then the answer might be different. For instance if we just ask whether every group can be embedded in $\operatorname{Aut}_k(V)$ for some (possibly infinite-dimensional) vector space over a field $k$, then the answer is clearly yes, over any field $k$: this follows from Cayley's theorem upon identifying permutations of $G$ with linear maps of the $k$-vector space with basis $G$.

-
The existence of a paper on counter-linear groups is not a proof of existence of counter-linear groups. :) –  Phira Nov 24 '11 at 20:33
"This paper" is Berrick, 1994 ams.org/mathscinet-getitem?mr=1285653 with several corrections in dx.doi.org/10.1017/S0004972700014787 that remove one of the examples. However, several examples are given in ams.org/mathscinet-getitem?mr=1277521 and ams.org/mathscinet-getitem?mr=981847 –  Jack Schmidt Nov 24 '11 at 20:58

Finitely generated linear groups are residually finite. Although this is a deep result, it allows you to conjure up non-linear groups willy-nilly.

For example, groups which are non-Hopfian are not linear. A good example of such a group is the Baumslag-Solitar group $$BS(2, 3)\cong \langle a, t; t^{-1}a^2t=a^3\rangle.$$ This group is not Hopfian (that is to say, there exists a surjective endomorphism which is not injective). For a proof, see Magnus, Karrass and Solitar, "Combinatorial Group Theory". It is well-known, and can be found in your favourite graduate group-theory text, or in this Math.SE answer, that finitely generated, residually finite groups are Hopfian. Thus, finitely generated, non-Hopfian groups are not residually finite and so not linear.

The question you are now asking is, of course, "are all finitely generated residually finite groups linear?". Well...no. There is a paper of Cornelia Drutu and Mark Sapir, where they prove that the group $$\langle a, t; a^{(t^2)}=a^2\rangle$$ is residually finite but non-linear.

-