What is known about the subgroup of $GL(n, \mathbb Z)$ fixing (under left multiplication) the vector ${(1, 1, \cdots, 1)}^T$ ('T' denotes transposition). I'm particularly interested in the case $n = 5$.
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5$\begingroup$ It is conugate to the subgroup fixing $(1,0,\dots,0)^T$ (because these two vectors belong to the same orbit), and therefore isomorphic to $\mathrm{GL}(n-1,\mathbf{Z}) \rtimes \mathbf{Z}^{n-1}$. $\endgroup$– Mikael de la SalleNov 25, 2020 at 15:48
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2$\begingroup$ The question is somewhat overly general. It is GL$(n,Z)$-conjugate (and if $ n \geq 3$, SL$(n,Z)$-conjugate) to the direct sum of $1$ and a matrix in $GL(n-1,Z)$ (complete the singleton set consisting of the vector to a $Z$.-basis), and every such is conjugate to one of these. What did you want to know? $\endgroup$– David HandelmanNov 25, 2020 at 15:49
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1$\begingroup$ I misread the question: I thought it was a single matrix, which is what the first "It" refers to. But the argument still works for all elements of the subgroup (since we can take the same basis). $\endgroup$– David HandelmanNov 25, 2020 at 15:58
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$\begingroup$ Thanks! The comments are informative enough. $\endgroup$– A. GuptaNov 25, 2020 at 16:09
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$\begingroup$ @MikaeldelaSalle: maybe isomorphic to ${\bf Z}^{n-1}\rtimes {\rm GL}(n-1, {\bf Z})$? I would say that the subgroup ${\bf Z}^{n-1}$ is a normal, while $ {\rm GL}(n-1, {\bf Z})$ is not. $\endgroup$– Mikhail BorovoiNov 26, 2020 at 6:46
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