Let \(H=G/K\) be the quotient of an abelian group \(G\) by a subgroup \(K\). Prove or disprove:
- If \(H\) is finite cyclic, then \(G\) is isomorphic to the direct product \(H\times K\).
- If \(H\) is a direct product of infinite cyclic groups, then \(G\) is isomorphic to \(H\times K\).
Solution
Let \[ q:G\longrightarrow H=G/K \] denote the quotient homomorphism.
1. Statement (1) is false.
Take \[ G=\ZZ/4\ZZ \] and let \[ K=2\ZZ/4\ZZ. \] Then \(K\cong\ZZ/2\ZZ\), and \[ H=G/K\cong\ZZ/2\ZZ, \] so \(H\) is finite cyclic. However, \[ H\times K \cong (\ZZ/2\ZZ)\times(\ZZ/2\ZZ) \] has no element of order \(4\), whereas \(G=\ZZ/4\ZZ\) does. Hence \(G\not\cong H\times K\).
2. If the direct product in statement (2) has finitely many factors, write\[ H=C_1\times\cdots\times C_r, \]where each \(C_i\) is infinite cyclic, and choose a generator \(h_i\) of \(C_i\).Then \(H\) is free abelian on \(h_1,\ldots,h_r\).
For a finite product, every element of \(H\) has a unique expression \[ n_1h_1+\cdots+n_rh_r \] with \(n_1,\ldots,n_r\in\ZZ\). Thus \(H\cong\ZZ^r\) with basis \(h_1,\ldots,h_r\).
3. Under the finite-product interpretation, the quotient map \(q\) admits ahomomorphic section\[ s:H\longrightarrow G \]with \(q\circ s=\operatorname{id}_H\).
For each \(i\in\{1,\ldots,r\}\), choose \(g_i\in G\) with \[ q(g_i)=h_i. \] Since \(H\) is free abelian on \(h_1,\ldots,h_r\) by step 2, there is a unique homomorphism \(s:H\to G\) satisfying \[ s(h_i)=g_i \qquad(1\leq i\leq r). \] For every \(i\in\{1,\ldots,r\}\), \[ (q\circ s)(h_i)=q(g_i)=h_i. \] Because the \(h_i\) generate \(H\), it follows that \(q\circ s=\operatorname{id}_H\).
4. The map\[ \Phi:K\times H\longrightarrow G, \qquad \Phi(k,h)=k+s(h), \]is an isomorphism.
It is a homomorphism because \(G\) is abelian. For \(g\in G\), put \[ h=q(g). \] Then \[ q\bigl(g-s(h)\bigr) =q(g)-(q\circ s)(h) =h-h =0, \] so \(g-s(h)\in K\). Hence \[ g=\Phi\bigl(g-s(h),h\bigr), \] and \(\Phi\) is surjective.
If \(\Phi(k,h)=0\), applying \(q\) gives \[ 0=q(k)+(q\circ s)(h)=h, \] because \(q(k)=0\) for \(k\in K\). Thus \(h=0\), and then \(k=\Phi(k,0)=0\). Therefore \(\Phi\) is injective.
5. Under the finite-product interpretation, statement (2) is true:\[ G\cong H\times K. \]
By step 4, \[ G\cong K\times H. \] Since direct products of abelian groups are symmetric, \(K\times H\cong H\times K\).
6. If arbitrary infinite direct products are allowed in statement (2),then statement (2) is false in general.
Take \(H=\prod_{n\geq1}\ZZ\). By the Baer–Specker theorem, \(H\) is not a free abelian group. Let \(F\) be the free abelian group on the underlying set of \(H\), and let \[ \pi:F\longrightarrow H \] send the basis element indexed by \(h\in H\) to \(h\). Then \(\pi\) is surjective. Put \(K=\ker\pi\). By the first isomorphism theorem, \[ F/K\cong H, \] so this is an instance of the quotient in the problem.
If \(F\cong H\times K\), then \(H\) is isomorphic to a direct summand, hence to a subgroup, of the free abelian group \(F\). Every subgroup of a free abelian group is free, contradicting the Baer–Specker theorem. Thus this quotient need not split when an arbitrary infinite product is permitted.
7. The conclusions are\[ \boxed{ \begin{aligned} &\text{(1) is false;}\\ &\text{(2) is true for a finite direct product of infinite cyclic groups,}\\ &\text{and false in general for arbitrary infinite direct products.} \end{aligned} } \]
Step 1 disproves statement (1). Step 5 proves statement (2) for the finite-product reading, and step 6 supplies the infinite-product counterexample.
8. Q.E.D.
Step 7 records the required conclusions.