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For fields $K$ and $L$, I am interested in proving that "$GL_n(K)$ and $GL_m(L)$ are isomorphic (as groups) if and only if $m=n$ and $K\simeq L$".

I don't know how generally this is true, but:

  • assume $K=L$. In that case, if $char(K) \neq 2$ then there is a usual proof that $n=m$ by looking at the group of involutions of each, that yields $2^m = 2^n$ and this the result.
  • if $char(K) = 2$, I do not find any proof of this fact.
  • is there a stronger result without supposing $K = L$? Can we at least conclude that $char(K) = char(L)$?
Eric Wofsey
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Desiderius Severus
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  • By "equivalent" you mean isomorphic as groups? – Eric Wofsey Apr 26 '20 at 00:54
  • @EricWofsey Yes, sorry I don't know how I wrote "equivalent" instead. I just edited my post – Desiderius Severus Apr 26 '20 at 01:00
  • There is a notion of equivalent linear groups, (which was stated previously by the OP) where the following statement holds true : "The two linear groups are equivalent if and only if n=m and K and L are elementary equivalent" ( this holds even if K and L are rings ). This can be extended to show that if K and L are algebraically closed, then char(K)=char(L). – Yuzuriha Inori Apr 26 '20 at 01:08
  • @YuzurihaInori Could you precise what is this equivalence notion? This is already a statement of interest, however I am still interested in the "isomorphic" version of the question – Desiderius Severus Apr 26 '20 at 01:23
  • I would suggest you take a look here https://www.math.upenn.edu/~pop/Research/files-Res/AWS03-fin.pdf . It is a concept borrowed from logic and applied to classify several algebraic objects. It can be shown that for finite objects, isomorphism and elementary equivalence are the same ( if I am not wrong, any two isomorphic objects are elementary equivalent but the other way around is generally not true ). – Yuzuriha Inori Apr 26 '20 at 01:30
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    you might find this helpful: https://mathoverflow.net/questions/106838/isomorphic-general-linear-groups-implies-isomorphic-fields – anonymous67 Apr 26 '20 at 05:54
  • @Learning Thanks a lot for the link, it is good to know the answer is essentially yes and generally. What about an accessible proof for, say, undergraduate? – Desiderius Severus Apr 26 '20 at 13:53

1 Answers1

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I think I have an answer for the case of $\mathrm{char}(K) = 2$ (when $K = L$).

First, for the case $\mathrm{char}(K) \neq 2$, I don't understand what you mean about the group of involutions in $GL_{n}(K)$. My interpretation is that if you look at the involutions of $GL_{n}(K)$ (i.e. the elements with $x^2 = 1$), they fall into one of exactly $n+1$ conjugacy classes: when $\mathrm{char}(K) \neq 2$, a matrix with $x^2 = 1$ is diagonalizable with eigenvalues $\pm 1$, hence the conjugacy class only depends on the multiplicity of the eigenvalue $-1$, which can be $0, 1, 2, \dots, n$. Thus we can distinguish the groups $GL_n(K)$ by looking at the number of conjugacy classes of involutions.

Let's try the same thing when $\mathrm{char}(K) = 2$. How many conjugacy classes of involutions are there in this case? We can write the condition $x^2 = 1$ as $(x - 1)^2 = 0$. A matrix satisfying this equation can be written in Jordan canonical form as a block matrix with Jordan blocks $$\begin{pmatrix} 1 \end{pmatrix} \,\,\text{and}\,\, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$ and the conjugacy class is determined by the numbers of each type of Jordan block, so the number of conjugacy classes is just the number of partitions of the number $n$ into parts of size $1$ or $2$, which is $\lfloor n/2 \rfloor + 1$. This almost works, but it fails to distinguish the case $n = 2k$ from $n = 2k+1$.

Instead, consider the number of conjugacy classes of elements which satisfy $x^4 = 1$. In characteristic $2$ this condition can be written as $(x - 1)^4 = 0$, so a matrix satisfying this equation can be written in Jordan canonical form as a block matrix with Jordan blocks $$\begin{pmatrix} 1 \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ where again the conjugacy class is determined by the numbers of each type of Jordan block. It follows that the number of conjugacy classes is the number $p_4(n)$ of partitions of the number $n$ into parts of size $1$, $2$, $3$, and $4$.

With some work, we can explicitly calculate this number as a function of $n$, but it suffices to show that it's strictly increasing in $n$. Note that for $n \geq 1$ we have an injection from the partitions of $n$ of this form to the partitions of $n+1$ of this form given by adding an additional part of size $1$. But we can represent all numbers $m \geq 2$ as a sum $2a + 3b$ for $a, b \geq 0$, so in particular, there is a partition of $n+1$ into only parts of size $2$ and $3$ -- which is necessarily not in the image of our injection. It follows that $p_4(n+1) \geq p_4(n) + 1$, so $p_4$ is strictly increasing. Then for $n \neq m$, $GL_n(K)$ and $GL_m(K)$ have distinct numbers of conjugacy classes of elements with $x^4 = 1$, since $p_4(n) \neq p_4(m)$, so they are not isomorphic.

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