Exercise N11.1 (1). Let be a map of sets. A function is called the left (resp. right) inverse of if (resp. ).
(a) Show that admits a right inverse if and only if is onto. Show that in general a right inverse is not unique.
(b) Show that admits a left inverse if and only if is one-to-one.
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(a) We divide into and .
CASE 1: there is with . Since for all there is an with , let for such an . Then in we can swap for , so we get , which we saw was .
CASE 2: for all . But note must be some , so really shows that for all there is an with , which is the definition of surjectivity.
(b) This is once again divided into two cases:
CASE: …
Exercise N11.2. Let be a group and let . Show that
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is not abelian (if it were, this would be trivial). Say ; well, and exists. must be some product in .
Need a different approach!!
Assume that did not hold. Additionally, let be a non-abelian group. Then . In order to guarantee that this results in , it must be abelian: does not necessarily equal unless each pair of ’s are next to each other.
Show ?
By induction. Base: ✓.
Assume it holds for ; we show it for . So and . Now, by associativity,
Well, and (with ), so this is . So is the inverse of .
Exercise N11.3 (5). Show that a group cannot be the union of two proper subgroups.
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Plan for #5: show that there must exist one element without an inverse.
Examples tried: ,,, so . Since , is not a proper subgroup?
: uses different operations, so it does not count. In the inverses are respectively. Trying and : the union .