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?

  • $\begingroup$ 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. $\endgroup$
    – davidlowryduda
    Commented Nov 24, 2011 at 18:00
  • $\begingroup$ 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 $F$. $\endgroup$ Commented Nov 25, 2011 at 11:26
  • $\begingroup$ 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. $\endgroup$
    – arsmath
    Commented Feb 12, 2014 at 12:27

5 Answers 5


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.

  • 11
    $\begingroup$ 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? $\endgroup$ Commented Nov 24, 2011 at 18:20
  • 1
    $\begingroup$ @QiaochuYuan Yes, this is true. $\endgroup$ Commented Dec 21, 2017 at 16:49
  • $\begingroup$ @Michael Slone: Isn't a finite group isomorphic to its regular representation ? It seems more intuitive to me as a proof of your first statement. $\endgroup$
    – Weier
    Commented Aug 28, 2020 at 10:15

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.

  • $\begingroup$ Just curious, can you elaborate on the second last sentence a little more? $\endgroup$ Commented Nov 24, 2011 at 21:05
  • $\begingroup$ @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. $\endgroup$ Commented Nov 25, 2011 at 1:37
  • $\begingroup$ I know that, but what does $\mathbb{T}^n$ mean here? $\endgroup$ Commented Nov 25, 2011 at 7:38
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    $\begingroup$ @Eric: I believe it means the $n$-dimensional torus. $\endgroup$ Commented Nov 26, 2011 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.

  • 3
    $\begingroup$ 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/…). $\endgroup$ Commented Nov 24, 2011 at 18:18
  • 4
    $\begingroup$ @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. $\endgroup$ Commented Nov 24, 2011 at 18:21
  • $\begingroup$ 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. $\endgroup$ Commented Nov 24, 2011 at 18:50
  • $\begingroup$ @Jason: is it obvious that you get every cardinality that way? $\endgroup$ Commented Nov 24, 2011 at 19:04
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    $\begingroup$ @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. $\endgroup$ Commented Nov 24, 2011 at 19:15

Finitely generated linear groups are residually finite. This is a deep result, due to Mal'cev (see here for a sketch proof), and it allows you to conjure up non-linear groups willy-nilly.

For example, the Baumslag-Solitar group $$BS(2, 3)\cong \langle a, t\mid t^{-1}a^2t=a^3\rangle$$ is non-residually finite, and hence non-linear. The proof that it is non-residually finite is not hard, and comes from combining the following two results. A group $G$ is Hopfian is every surjective endomorphism $G\twoheadrightarrow G$ is injective.

  1. $BS(2, 3)$ is non-Hopfian, as the map $a\mapsto a^2, t\mapsto t$ extends to a surjective but non-injective homomorphicm (see here).
  2. Residually finite groups are Hopfian (see here).

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 $$\langle a, t\mid a^{(t^2)}=a^2\rangle$$ is residually finite but non-linear.


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 $G$.


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