I Mistake in Schaum's Group Theory?

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fresh_42

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If I understand what your notation conveys, the last bit should be ## \stackrel{+\frac{1}{4}}{\hookrightarrow} (0,1)##, since you're adding 1/4 to each element of the punctured interval (-1/4, 1/4).
No, I just wanted to shift it into the interval injectively. The amount didn't matter. With ##\frac{1}{4}## I would have had to deal with the zero, which I avoided by taking ##\frac{1}{3}##.
 

WWGD

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We still have the difficulty that zero divisors are possible as long as we haven't a group. The (unique) solvability of ##x A = B## is equivalent to the group axioms, so we circle around the main topic. Guess we have to visit La La Land.
Where are you getting your scalars? We're talking coefficients in a field of characteristic zero now, aren't we?
 

fresh_42

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Where are you getting your scalars? We're talking coefficients in a field of characteristic zero now, aren't we?
O.k., but matrices can still multiply to zero. However, since @mathwonk #49 I consider our riddle solved.
 

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