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Letting $\mathbb{F}_{1}$ and $\mathbb{F}_{2}$ be fields, and letting $n \geq 3$ and $m$ be natural numbers, it is known that $\text{GL}_{m}(\mathbb{F}_{1})$ and $\text{GL}_{n}(\mathbb{F}_{2})$ are elementarily equivalent if and only if $m=n$ and $\mathbb{F}_{1} \equiv \mathbb{F}_{2}$ (as proven in "Elementary Properties of Linear Groups" in the collection "The Metamathematics of Algebraic Systems — Collected Papers: 1936–1967").

So, given a field $\mathbb{F}$, if $n \neq m$, then $\text{GL}_{m}(\mathbb{F}) \not\equiv \text{GL}_{n}(\mathbb{F})$, and thus $\text{GL}_{m}(\mathbb{F}) \not\cong \text{GL}_{n}(\mathbb{F})$.

Letting $R$ be a commutative ring (with unity), and letting $n, m \in \mathbb{N}$ be such that $n \neq m$, is there a simple "algebraic" way of proving that $\text{GL}_{m}(R)$ and $\text{GL}_{n}(R)$ are not isomorphic (as groups)? Is there a simple group-theoretic way of showing that $\text{GL}_{m}(\mathbb{F}) \not\cong \text{GL}_{n}(\mathbb{F})$ for a field $\mathbb{F}$?

Certain special cases of this problem trivially hold, for example in the case whereby $\mathbb{F}$ is finite, in which case $|\text{GL}_{m}(\mathbb{F})| \neq |\text{GL}_{n}(\mathbb{F})|$.

John M. Campbell
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  • $n=1$ is distinguished by the fact that $\text{GL}n(R)$ is abelian. For $n>1$ and $R=\mathbb{F}$ a field, then $n$ is the largest integer $m$ such that $\text{GL}_n(\mathbb{F})$ has a subgroup isomorphic to the symmetric group $S{m+1}$. Is that the kind of characterization you're looking for? This may also work for arbitrary $R$ (it certainly does for integral domains), but I don't know what's known about representations of symmetric groups over general commutative rings. – Jeremy Rickard Mar 07 '16 at 10:32
  • @JeremyRickard Thank you for your response. The characterization given above is exactly the kind of characterization I was looking for. Is there a well-known representation-theoretic proof that the largest natural number $m$ such that $\text{GL}{n}(\mathbb{F})$ has a subgroup isomorphic to $S{m+1}$ is $n$ (for $n > 1$)? – John M. Campbell Mar 07 '16 at 11:36
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    @JohnM.Campbell Actually, what I claimed is slightly wrong. If the characteristic of $\mathbb{F}$ divides $n+2$ then $\text{GL}n(\mathbb{F})$ has a subgroup isomorphic to $S{n+2}$. So this fact won't distinguish between $\text{GL}{kp-2}$ and $\text{GL}{kp-1}$ if $\text{char}(\mathbb{F})=p>0$. – Jeremy Rickard Mar 07 '16 at 11:57
  • @ John M. Campbell That $GL_{m}(F) \not \cong GL_{n}(F)$ for $m \neq n$ is right? – bing Aug 30 '16 at 03:46
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    There is a different but somehow similar question here: https://math.stackexchange.com/questions/814100. Moreover, this question on MO looks interesting. – Watson Nov 28 '16 at 20:43

2 Answers2

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If the ring is a field $F$ with characteristic not equal to $2$, then it is not hard to prove that the largest $k$ with $C_2^k \le {\rm GL}(n,F)$ is $k=n$.

We prove this by induction on $n$. Since $-1$ is the only element of order $2$ in $F$, the result holds for $n=1$, so assume that $n>1$. Then $k>1$, so we can choose an element $x$ in this subgroup with $x \ne -I_n$. Then $x$ is conjugate to a diagonal matrix with $1$s and $-1$s on the diagonal, and its centralizer is isomorphic to ${\rm GL}(m,F) \times {\rm GL}(n-m,F)$ for some $m$ with $1 \le m < n$. Now the result follows by induction.

I don't know whether there is an equally elementary argument for fields of characteristic $2$.

Derek Holt
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  • I didn't get the conclusion of the proof. I mean how showing $C_2^n \leq \text{GL}(n,F)$ is enough to show that $\text{GL}(n,F) \not\cong \text{GL}(m,F)$ ? Can you please explain it ? – MAS Sep 16 '20 at 13:29
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    $C_2^n \le {\rm GL}(n,F)$ is easy to show. What I showed in addition is that $C_2^m \not\le {\rm GL}(n,F)$ when $m > n$. – Derek Holt Sep 16 '20 at 17:56
  • Thanks very much – MAS Sep 16 '20 at 18:27
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Let $G=GL_n(K)$ where $K$ is a field of characteristic not 2. Let $A$ be a maximal subgroup of $G$ of exponent 2. As every element of order 2 of $GL_n(K)$ is diagonalizable (with $1$ or $-1$ as eigen-values) and since $A$ is abelian, elements of $A$ are simultaneously diagonalizable. Hence we may assume that $A$ consists of matrices with 1's and $-1$'s on the diagonal. Therefore $A$ is isomorphic to the direct sum of $n$ copies of $\{1, -1\}$ and hence has order $2^{n}$. This shows that $GL_n(K)$ determines $n$. The same argument works for $SL$ instead of $GL$ if $n \geq 2$. (I just noticed that this same answer but with a different argument was given above.)

Here is a question: Assume characteristics are not 2. I can show in a quite elementary way that if the statement $SL_2(K) \simeq SL_2(L) \implies K \simeq L$ holds, then for $n \geq 2$, the statement $SL_n(K) \simeq SL_n(L) \implies K \simeq L$ holds. But I do not know how to prove this for $n=2$ in its full generality. We can of course assume the groups i.e. the fields are infinite.

On the other hand by using any non-central diagonal element as a parameter, one can define the field $K$ in the group $SL_2(K)$ as follows. Let $t_0$ be one such element. Let $T=C_{SL_2(K)}(t_0) \simeq K^*$ (torus). We may regard $T$ as the group diagonal matrices of determinant 1. There are exactly two subgroups of $SL_2(K)$ of the form $\langle u^T\cup\{1\} \rangle$ for any $1\neq u$ in the subgroup, the strictly upper and lower triangular matrices, say $U$ and $V$ (unipotent) respectively. (Because $x = (1+x/2)^2 - 1^2 - (x/2)^2$ for any $x\in K$, see below). They are both isomorphic to the addive group of $K$. Choose one of them, say $U$. Denote the elements of $T$ by $t(x)$ where $x\in K^*$ and elements of $U$ by $u(y)$ where $y\in K$. Then $T$ acts on $U$ as follows $u(y)^{t(x)} = u(x^2y)$. Thus we get the subfield of $K$ generated by the squares. But since $x = (1+x/2)^2 - 1^2 - (x/2)^2$ for any $x\in K$, the subfield generated by the squares is $K$ itself. Thus the field $K$ is definable with one parameter, namely $t_0$. (Except that the group does not know the unit element 1 of the field, we only get an affine version of a field, something like $K$ with addition and a ternary multiplication $xy^{-1}z$; to fix 1 of the field $K$ we need one more parameter, but this is irrelevant to us). It follows that in the group $SL_2(L)$ both fields $K$ and $L$ are definable.

In particular if the automorphism takes a non-central diagonalizable element of $SL_2(K)$ to a non-central diagonalizable element of $SL_2(L)$, then we will necessarily have $K\simeq L$. This will be so if we can distinguish diagonalizable elements of $SL_2(K)$ from its non-diagonalizable semisimple elements (i.e. diagonalizable in the algebraic closure) in a group theoretic way.

Ali Nesin
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