Let $U=U(n,R)$ the group of $n\times n$ matrices over R (a commutative unitary ring) with $1$ on the diagonal and $0$ under it. For each $0<i\leq n$ let $U_i$ be the subgroup of $U$ of all elements of $U$ where the first $i-1$ superdiagonals are zero. This way we have that $1=U_n<\ldots<U_1=U$. It is easily shown by computation that this is a central series of $U$, which is hence a nilpotent group. Is it true that $U/U_i$ is isomorphic with $U(n-i,R)$? I'm not used to work with matrices, so I am not satisfied with any approach I try to enact.

Would you give me a formal, maybe even "nice" description of this isomorphism, if there is one?

  • $\begingroup$ I'm trying to work out a simple case, $n=2, i=1$. If I understand the setup, $U_i $ is the trivial subgroup. The claim would then be $U=U(2,R)$ is isomorphic to $U(2-1,R)$. This is false for (say) $R$ the integers (or the real numbers, if you prefer). $\endgroup$
    – hardmath
    May 6 '18 at 23:18
  • $\begingroup$ You're right. $U_i$ is the subgroup of elements with the first $i+1$ superdiagonals equal to zero. So, if $n=2$ and $i=1$ you have that $U/U_1=1=U(1,R)$. Now I've corrected the typo. $\endgroup$
    – Alex Doe
    May 7 '18 at 6:44
  • 1
    $\begingroup$ Your change has made it even worse! You say "it is known that". Where does this come from? $\endgroup$
    – Derek Holt
    May 7 '18 at 7:25
  • $\begingroup$ I was sure of having read it in one of Fuchs' books as an exercise or something similar, but now I'm starting to doubt it... Maybe you should read the question as (or it should be changed into) a "Is it true that...?" one. $\endgroup$
    – Alex Doe
    May 7 '18 at 7:47
  • $\begingroup$ I've corrected my question, changing its nature and adding a bit more of an explanation. I hope this will be all right and that those groups are really isomorphic :) $\endgroup$
    – Alex Doe
    May 7 '18 at 21:10

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