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Apr 6, 2013 at 17:04 comment added Alireza Abdollahi @RDK: From where you know $x^{2^d}$ is non-trivial? Note that the relation $x^{2^{d+1}}=1$ cannot solely imply that the order of $x$ is $2^{d+1}$. The answer of Derek Holt can help you to understand the main difficulty.
Apr 6, 2013 at 6:44 history answered Soluble CC BY-SA 3.0