Let $A\in GL_n(\mathbb F_2)$ be an element of order greater than or equal to $2^n-1$ . Then is it true that order of $A$ is $2^n-1$ ?
I know that $|GL_n(\mathbb F_2)|=(2^n-1)(2^n-2^2)...(2^n-2^{n-1})$. But I'm unable to proceed further.
Please help.
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1Do you know how to get an element of exact order $2^n-1$ ? – Lubin Oct 15 '18 at 04:07
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3See https://math.stackexchange.com/questions/292620/exponent-of-gln-q and https://mathoverflow.net/questions/109483/maximal-order-of-elements-in-gln-p/109488#109488 – Gerry Myerson Oct 15 '18 at 04:15
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1@GerryMyerson: thanks for the links ... I've got it now ... – user521337 Oct 15 '18 at 07:15
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1Maybe you could write up and post an answer, if you don't think it's a duplicate of what's in the stackexchange link. – Gerry Myerson Oct 15 '18 at 07:45
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Maybe you could do it today. – Gerry Myerson Oct 17 '18 at 00:55
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@GerryMyerson: oh no well it is really a duplicate, you should mark it as a duplicate – user521337 Oct 17 '18 at 01:08