# I Mistake in Schaum's Group Theory?

#### fresh_42

Mentor
2018 Award
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

Gold Member
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

Mentor
2018 Award
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.