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I have 2 groups:

  • general linear $ k \times k $ with $\cdot$
  • top-triangle matrix $ n \times n $ with 1 on main diagonal. Operation is $\cdot$ too

Is there isomorphism for any any non-trivial $n,k$ i.e $n \neq 2 \ or \ k \neq 1$ over $\mathbb{R}$ or $\mathbb{Q}$?

If no, how can I prove it?

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Did you intend $\mathbb{N}$ to be the set of integers? natural numbers? –  Bill Cook Dec 15 '11 at 13:20
    
@BillCook, I mixed it up with $\mathbb{Q}$, fixed now –  RiaD Dec 15 '11 at 13:21
2  
Upper-triangular matrices form solvable groups, general linear groups are not solvable (for $k>1$). Thus they cannot be isomorphic. –  Bill Cook Dec 15 '11 at 13:22
    
@BillCook, Hmm, ok. Can you post as an answer? I'll accept it –  RiaD Dec 15 '11 at 13:27

2 Answers 2

up vote 6 down vote accepted

Upper-triangular matrices form solvable groups, general linear groups are not solvable (for $k>1$). Thus they cannot be isomorphic.

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Upper triangular matrix groups with 1 on the main diagonal are even nilpotent. –  Jack Schmidt Dec 16 '11 at 12:28

Another reason is that if the field is of characteristics $0$ then all elements (except the identity matrix) in the set upper triangle matrix with 1 on the main diagonal do not have finite order. However, there are lots of matrix in $ GL_k(F)$ has finite order. For instance, those have $-1$ or $1$ on the main diagonal and $0$ elsewhere.

However, this argument doesn't work for fields of finite characteristics.

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