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Sep 24, 2017 at 3:20 comment added Dianbin Bao Dear Bogdanov, thanks for your comment, I see the mistake in my argument.
Sep 23, 2017 at 18:56 comment added Ilya Bogdanov Notice that $\mathbb Q[\lambda_1]\neq\mathbb Q[\lambda_2]$, so you cannot take their images under $\phi$ simultaneously --- or, if you define it exactly as you do, you will have $\phi(\lambda_1)=\phi(\lambda_2)$.
Sep 23, 2017 at 17:08 comment added Dianbin Bao $\phi(\lambda_1)\phi(\lambda_2)=\phi(\lambda_2)\phi(\lambda_1)$ since $\lambda_1\lambda_2=\lambda_2\lambda_1$. The three matrices commute.
Sep 23, 2017 at 17:05 history edited Dianbin Bao CC BY-SA 3.0
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Sep 23, 2017 at 16:54 comment added Dianbin Bao You can use the fact that $\lambda_1\lambda_2\lambda_3=1$ to get $\phi(\lambda_1)\phi(\lambda_2)\phi(\lambda_3)$ is the identities map. Since you also have $\lambda_1\lambda_3\lambda_2=1$, you can switch order. Then the three matrices $\phi(\lambda_i)$ work.
Sep 23, 2017 at 16:48 history edited Dianbin Bao CC BY-SA 3.0
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Sep 23, 2017 at 16:48 comment added Misha OK, you found one NRS matrix; how do you find three NRS matrices $A, B, C$ whose product is $1$ and which generate a subgroup isomorphic to ${\mathbb Z}^2$?
Sep 23, 2017 at 16:43 history answered Dianbin Bao CC BY-SA 3.0