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I discovered the same conjectur: U(r) * U(s) = U(r*s)

Revision as of 22:25, 4 February 2009


For this problem, does anyone know what the new conjecture is supposed to be? I thought it might just be that the order of r multiplied by the order of s is NOT the order of rs, but I wasn't sure if there was another conjecture that could be made. --Clwarner 21:14, 3 February 2009 (UTC)


The conjecture I made was that $ \scriptstyle gcd(r,s)=1\ \leftrightarrow\ \mid U(r)\mid*\mid U(s)\mid=\mid U(rs)\mid $.

--Nick Rupley 05:01, 4 February 2009 (UTC)

I made an equivalent conjecture: If r and s are relatively prime, then U(r) * U(s) = U(r*s).

--Yu Suo

I discovered the same conjectur: U(r) * U(s) = U(r*s)

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

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