Hatcher Exercise 2.1.14

(a) Consider . We wish to determine if there is a choice of and that makes this into a short exact sequence.

Indeed, pick to be the map sending . Then, . Notice then that is group of order 4. One can check that the element has order 4, so it must generate the group and the quotient is isomorphic to . This lets us define as the quotient map, giving a short exact sequence.

(b) Now consider , for prime. We wish to find which abelian groups admit maps that make the sequence exact. First note that, for this sequence to be exact, we need . Thus, must have order .

A certain class of groups of order can be constructed as follows: for . Notice when is minimal we get and when is maximal we get (or the other way around).

The maps can be explicitly constructed. For , let be the map sending . Then with (which is of order ), we must have . We claim that is an element of order , and thus the quotient is isomorphic to . To see this, consider an arbitrary coset in represented by . Notice that represents the same coset, since we are subtracting a multiple of an element of . But, we have:

What we’ve shown is that every coset, and therefore element of , is represented by a multiple of . So, generates the (cyclic) quotient group, so it must be isomorphic to . Then, let be the quotient map and we have a short exact sequence.

So all of the groups are acceptable. To show that these are all the groups (up to isomorphism, of course), we note that is generated by two (non-unique) elements: and such that and (this implies that must be isomorphic to direct product of two cyclic groups, since and necessarily generate cyclic subgroups):

To see this, let be arbitrary. There are two possibilities: either or not.

Suppose . Then is a multiple of (since is a cyclic subgroup).

Suppose . Then we can also conclude that . This means that is a non-trivial element, and thus a multiple of , i.e. . This means that , so that . Then, , so we have . We now conclude that for some . So in both cases, is expressed as a combination of and .

(c) Finally, consider .

Here, we show explicitly that for all such that are acceptable. Set . Let be the map

Let be the map

Since , we have .

Let . Then, and . Then, (modulo ). So, . So we have

Then , and with this choice of and makes the sequence exact.