User john mac - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-21T13:02:48Zhttp://mathoverflow.net/feeds/user/5900http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/18271/what-out-of-print-books-would-you-like-to-see-re-printed/74619#74619Answer by John mac for What out-of-print books would you like to see re-printed?John mac2011-09-05T20:44:45Z2012-02-19T15:03:23Z<p>Perhaps we should be asking why excellent books are out of print. By what mechanisms can they be brought back to life? Can we learn from any successful campaigns?</p>
<p>For me: Lectures on the theory of functions of a complex variable Vols I & II
published in the 1960's by Noordhoff. A beautiful book authored by Sansone &
Gerretsen.</p>
http://mathoverflow.net/questions/82176/groups-with-a-representation-of-degree-n-for-each-n-1-2-3Groups with a representation of degree n for each n = 1,2,3,...John mac2011-11-29T13:37:55Z2011-11-29T13:37:55Z
<p>The group SU2(C) has this property. What other groups share it?</p>
http://mathoverflow.net/questions/79883/galois-theory-and-algorithms/80132#80132Answer by John mac for Galois theory and algorithmsJohn mac2011-11-05T13:29:45Z2011-11-05T13:29:45Z<p>Hermite ca 1858 in Comptes Rendus, solved the general 5-ic using the j-function
and the modular polynomial Phi_5(j(z), j(5z)). </p>
http://mathoverflow.net/questions/80127/being-a-subgroup-proof-by-character-theory/80131#80131Answer by John mac for Being a subgroup: proof by character theoryJohn mac2011-11-05T13:21:38Z2011-11-05T13:21:38Z<p>More enticing is the question of determining the existence of a subgroup H (which need not be normal) of G from the character table of G. </p>
http://mathoverflow.net/questions/78424/has-anyone-seen-this-graph/78509#78509Answer by John mac for Has anyone seen this graph?John mac2011-10-18T23:56:17Z2011-10-18T23:56:17Z<p>Any connected trivalent graph realises a Schreier coset graph of a subgroup of
the modular group. This yields a transitive permutation group of degree 16 x 3
since we expand each node into an oriented triangle. In the original graph,
switching the ends of edges & rotating the oriented triangles provides
generators of order 2 and order 3. Recall PSL(2,Z) = free product C2 * C3.</p>