I Mistake in Schaum's Group Theory?

  • Thread starter jstrunk
  • Start date

fresh_42

Mentor
Insights Author
2018 Award
11,708
8,229
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

Science Advisor
Gold Member
4,769
2,111
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
Insights Author
2018 Award
11,708
8,229
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.
 

Want to reply to this thread?

"Mistake in Schaum's Group Theory?" You must log in or register to reply here.

Related Threads for: Mistake in Schaum's Group Theory?

  • Posted
Replies
6
Views
726
  • Posted
Replies
1
Views
1K
  • Posted
Replies
1
Views
2K
  • Posted
Replies
6
Views
2K
  • Posted
Replies
10
Views
3K
  • Posted
Replies
2
Views
3K
  • Posted
Replies
6
Views
2K
  • Posted
Replies
1
Views
1K
Top