#51

There is a significant difference between #51 and Leduc's example, namely in #51 they want

to be one-to-one and onto. This implies that

is not trivial. After that like in Leduc's example we can show

,

,

PS

is not a homomorphism as

#55

if

then there exists

such that

Thus we can easily eliminate (A), (D) (

), and (C), (B) (

) since they do not generate proper subgroups of

. (E) is obviously fine for

.

#56

This is a tricky question.

I. Is true since for all

such that every open set that contains them contains at least one other point in A, obviously have the same property for (

), thus

.

Conversely if

holds the property that every open set that contains it contains at least one other point from

, we can assume WLOG that all of those open sets hold a point from

, but this means that

, so

II. A nice counterexample is

,

III. Is true since

, which means that

is closed.

IV.

is counterexample.