Determine whether there exists a short exact sequence \(0 \to \mathbb{Z}_4 \to \mathbb{Z}_8 \oplus \mathbb{Z}_2 \to \mathbb{Z}_4 \to 0\). More generally, determine which abelian groups \(A\) fit into a short exact sequence \(0 \to \mathbb{Z}_{p^m} \to A \to \mathbb{Z}_{p^n} \to 0\) with \(p\) prime. What about the case of short exact sequences \(0 \to \mathbb{Z} \to A \to \mathbb{Z}_n \to 0\)?
Solution
1. There is a short exact sequence\[ 0\longrightarrow\mathbb Z_4 \longrightarrow\mathbb Z_8\oplus\mathbb Z_2 \longrightarrow\mathbb Z_4 \longrightarrow0. \]
Define \[ i(1)=(2,1) \] and \[ q(a,b)=a-2b\pmod4. \] The element \((2,1)\) has order \(4\), so \(i\) is injective. The map \(q\) is surjective since \(q(1,0)=1\). Its kernel has order \[ \frac{|\mathbb Z_8\oplus\mathbb Z_2|}{4}=4 \] and contains the order-\(4\) subgroup generated by \((2,1)\), hence \[ \ker q=\operatorname{im}i. \]
Now fix a prime \(p\) and positive integers \(m,n\).
2. Every extension\[ 0\to\mathbb Z_{p^m}\to A\to\mathbb Z_{p^n}\to0 \]has a presentation\[ A\cong \left\langle x,y\ \middle|\ p^m x=0,\ p^n y=cx\right\rangle \]for some integer \(c\), well-defined modulo \(p^m\) up to the changes arising from replacing the chosen lift \(y\).
Let \(x\) be the image of a generator of the kernel. Choose \(y\in A\) mapping to a generator of the quotient. Since the quotient class of \(y\) has order \(p^n\), one has \[ p^n y=cx \] for some \(c\). The kernel is generated by \(x\) with relation \(p^m x=0\), and \(x,y\) generate \(A\), giving the presentation.
3. Put\[ r=\min\{v_p(c),m,n\}, \]with \(v_p(0)=\infty\). Then\[ A\cong \mathbb Z_{p^r}\oplus\mathbb Z_{p^{m+n-r}}, \]where \(\mathbb Z_{p^0}\) denotes the trivial group.
The relation matrix for the presentation in 2 is \[ M= \begin{pmatrix} p^m&0\\-c&p^n\end{pmatrix}. \] The product of the Smith invariant factors is \[ |\det M|=p^{m+n}. \] The first invariant factor is the gcd of all entries of \(M\), namely \[ p^{\min(m,n,v_p(c))}=p^r. \] Hence the second invariant factor is \(p^{m+n-r}\), giving the stated decomposition.
4. Conversely, every value\[ 0\le r\le\min(m,n) \]occurs. Thus the possible middle groups are exactly\[ \boxed{ \mathbb Z_{p^r}\oplus\mathbb Z_{p^{m+n-r}} \qquad(0\le r\le\min(m,n)). } \]
For \(r<\min(m,n)\) use the presentation \[ \langle x,y\mid p^m x=0,\ p^n y=p^r x\rangle. \] For \(r=\min(m,n)\) take \(c=0\). In every case the subgroup generated by \(x\) has order exactly \(p^m\): if \(kx\) lies in the relation subgroup in \(\mathbb Zx\oplus\mathbb Zy\), comparison of \(y\)-coordinates forces the coefficient of the second relation to be zero, so \(p^m\mid k\). The quotient by \(\langle x\rangle\) is generated by the class of \(y\) with exact order \(p^n\). Hence these presentations really give the required short exact sequences.
For the final question consider \[ 0\to\mathbb Z\to A\to\mathbb Z_n\to0. \]
5. Every such \(A\) has a presentation\[ A\cong\langle x,y\mid ny=cx\rangle \]for some \(c\in\mathbb Z\).
Choose \(x\) generating the infinite cyclic kernel and choose a lift \(y\) of a generator of the quotient. Then \(ny\) lies in the kernel, so \(ny=cx\). As before, \(x\) and \(y\) generate \(A\).
6. If\[ d=\gcd(c,n), \]then\[ A\cong\mathbb Z\oplus\mathbb Z_d. \]Consequently the possible middle groups are exactly\[ \boxed{\mathbb Z\oplus\mathbb Z_d\qquad(d\mid n),} \]with \(\mathbb Z_1=0\).
The single relation vector is \((-c,n)\). Smith normal form replaces it by \((d,0)\), so \[ \mathbb Z^2/\langle(-c,n)\rangle \cong\mathbb Z\oplus\mathbb Z_d. \] Every divisor \(d\) of \(n\) occurs by taking \(c=d\) (and \(d=n\) can also be represented by \(c=0\)). The element \(x\) remains of infinite order, since a relation killing a nonzero multiple of \(x\) would have to have zero \(y\)-coordinate and hence zero coefficient of \((-c,n)\). Thus each presentation gives the required extension.