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Let $G$ be an arbitrary group (finite or infinite), and let $GL_n(\mathbb C)$ be the general liner group. And let $\varphi : G \rightarrow GL_n(\mathbb C)$ be a homomorphic map.
My question is:
Can we always find the homomorphic map $\varphi$ be injective? In another word, can any infinite group be represented by some $n \times n$ complex matrices, which are matched with a single element in $G$?
Above are some examples.
Finite group like Hamilton IV group, $\mathrm Q=\{\mathrm{1, i, j, k}\}$, can be represented by these 8 complex matrices,
where:
$1=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\mathrm{i}= \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix},\mathrm{j}= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},\mathrm{k}= \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}. $

And infinite group like $\mathbb Z^*$ can be simply represented by $1\times 1$ matrices:
$(e^n), n \in \mathbb Z.$
So this problem occured to me.

Now, one idea is that the infinite group $G$ is generated by finite elements $\{1, x_1, ..., x_n\}$, and these elements are somewhat 'isolated' from each other. The word 'isolated' may mean that $ x_m \not= x_1^{r_1}x_2^{r_2}\dots x_{m-1}^{r_{m-1}}.$ But at least now, there are still 3 bad things about the idea:
1. How to define the word 'isolated'?
2. How to solve the problem of non-abelian situations?
3. Some infinite groups such as $\mathbb R^*$ cannot be generated by a few elements, but it can still be represented by matrices.
user26857
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Anton LL
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  • I guess the question is : "can we always find a map $\varphi$ which is injective ?" – TheSilverDoe Mar 01 '22 at 16:39
  • Sorry, I got it wrong, it should be injective – Anton LL Mar 01 '22 at 16:41
  • ACTUALLY, the problem is about infinite groups, for finite groups can seem to be represented by subgroups of symmetric groups $S_n$ – Anton LL Mar 01 '22 at 16:46
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    Have you heard about representation theory ? Basically, you are asking : does every group admit a faithful representation. Indeed, the answer is yes for finite groups, and no for infinite groups. – TheSilverDoe Mar 01 '22 at 16:49

2 Answers2

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All finite groups are isomorphic to a subgroup of $Gl_n(\mathbb{C})$ for some $n$. This can be shown by first showing that all finite groups are subgroups of a permutation group and then that all permutation groups sit inside a matrix group.

Infinite groups in general need not be isomorphic to a subgroup of $Gl_n(\mathbb{C})$ for any $n$. If I recall correctly one of the smallest/ simplest counter examples consists of infinitely many copies of $\mathbb{Z}_2$. Essentially one can build infinite groups that are just too big to fit into $Gl_n(\mathbb{C})$.

quarague
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    Right. One can build groups $G$ such that there isn't even an injective function $\phi$ from $G$ into any $\mathop{GL}_n({\Bbb C})$ on cardinality grounds. – Jamie Radcliffe Mar 01 '22 at 16:51
  • THATS Right, I did ignore the problem of cardinality. But if the group's cardinality is proper? (p.s. I know little about the cardinality theorem) – Anton LL Mar 01 '22 at 17:03
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    For $(\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}}$, its not a size argument (it has the correct cardinality), but rather that $GL_n(\mathbb{C})$ can only have $2^n$ commuting elements of order $2$ (as they can be simultaneously diagonalized). – Captain Lama Mar 01 '22 at 17:04
  • So, besides the corresbonding quantity of the elements（I dont know how to describe the meaning, but you may understand it）, what else resrict the existence of the map? – Anton LL Mar 01 '22 at 17:10
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I'll write a stronger statement: if $G$ is an abelian group of finite exponent (so all the elements have order less than, say, $r$), then $G$ admits an embedding in some $GL_n(\mathbb{C})$ if and only if it is finite.

Indeed, if $G$ is finite, it has a faithful representation, like any finite group. And the issue when $G$ is infinite is that $GL_n(\mathbb{C})$ cannot have an abelian subgroup of exponent $r$ with more than $r^n$ elements, because such a subgroup can be simultaneously diagonalized, and is therefore, up to conjugation, a subgroup of $\mu_r(\mathbb{C})^n\subset D_n(\mathbb{C})\subset GL_n(\mathbb{C})$ (here $D_n(\mathbb{C})$ is the subgroup of diagonal matrices in $GL_n(\mathbb{C})$).

Captain Lama
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  • ... and I would add to your statement that you can have a majoration of the cardinality of $G$ that depends uniquely on its exponent. In particular, this has interesting consequences, such as the fact that for fixed $n$, there is only a finite number of subgroup of $GL_n(\mathbb{Q})$ (up to isomorphism). – TheSilverDoe Mar 01 '22 at 19:15