**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.