Take two nonisomorphic perfect groups of order 1344 and label the elements of each with the numbers 1 through 1344, then superimpose their respective Cayley tables (for simplicity’s sake, the nth row and nth column should correspond to element n). What is the lowest possible number of mismatched entries?
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4$\begingroup$ I am dying to know how this question arose! $\endgroup$– Steven LandsburgCommented Jun 17, 2023 at 3:56
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$\begingroup$ I was wondering about how similar two groups of the same “shape” really are. $\endgroup$– Daniel SebaldCommented Jun 17, 2023 at 4:05
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$\begingroup$ The only cardinal of a nonabelian simple group dividing $1344$ is $168$, and $1344=168\times 8$. Since the Out of a group of order 8 has order $<168$, the only possibility will be a central extension. But the Schur multiplier of the simple group of order 168 has order 2. So I don't see how a perfect group of order 1344 could exist. $\endgroup$– YCorCommented Jun 17, 2023 at 7:38
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1$\begingroup$ There are two isomorphism classes of perfect groups of order $1344$. They are the split and the nonsplit extensions of $L_3(2)$ by the natural module. $\endgroup$– Dave BensonCommented Jun 17, 2023 at 8:01
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1$\begingroup$ It is hard to envisage how this problem could be solved without a massive computer calculation, which at first sight requires checking all $1344!$ possible mappings. In the end you might have to settle for an approximate answer. $\endgroup$– Derek HoltCommented Jun 17, 2023 at 10:10
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