Suppose \(G\) is a finite group such that for every subgroup \(H\le G\) there is a homomorphism \[ \phi:G\to H \] with \(\phi(h)=h\) for every \(h\in H\). Show that \(G\) is a direct product of groups of prime order.
Solution
1. The claim holds when \(G\) is trivial.
The trivial group is the empty direct product of groups of prime order.
2. Assume \(G\) is nontrivial and that the claim holds for allsmaller finite groups satisfying the same retract hypothesis. Choosea subgroup \(H\le G\) of prime order \(p\).
By Cauchy’s theorem, if a prime \(p\) divides \(\abs G\), then \(G\) has an element of order \(p\). The subgroup it generates has order \(p\).
3. Let\[ \phi:G\longrightarrow H \]be a retraction supplied by the hypothesis, and set\[ K=\ker\phi. \]Then\[ \abs G=\abs K\,\abs H \]and \(K\) is strictly smaller than \(G\).
Since \(\phi\) restricts to the identity on \(H\), it is surjective. The first isomorphism theorem therefore gives \[ G/K\cong H, \] so \[ [G:K]=\abs H=p. \] Thus \(\abs G=\abs K\,\abs H\) and \(\abs K<\abs G\).
4. The group \(K\) satisfies the same retract hypothesis.
Let \(L\le K\). By the hypothesis on \(G\), there is a homomorphism \[ \rho:G\longrightarrow L \] such that \(\rho(\ell)=\ell\) for every \(\ell\in L\). Restricting \(\rho\) to \(K\) gives a homomorphism \[ \rho|_K:K\longrightarrow L \] which is still the identity on \(L\). Hence every subgroup of \(K\) is a retract of \(K\).
5. The group \(K\) is a direct product of groups of prime order.
By steps <1>3–<1>4, \(K\) is smaller than \(G\) and satisfies the induction hypothesis.
6. Let\[ \sigma:G\longrightarrow K \]be a retraction onto \(K\), and define\[ \alpha:G\longrightarrow K\times H, \qquad \alpha(g)=(\sigma(g),\phi(g)). \]Then \(\alpha\) is injective.
The subgroup \(K\le G\) has a retraction \(\sigma\) by the original hypothesis. Both components of \(\alpha\) are homomorphisms, so \(\alpha\) is a homomorphism.
If \(g\in\ker\alpha\), then \(\phi(g)=1\), so \(g\in K\). But \(\sigma|_K=\operatorname{id}_K\), while \(\sigma(g)=1\). Hence \(g=1\). Thus \(\ker\alpha=\{1\}\) and \(\alpha\) is injective.
7. The map \(\alpha\) is an isomorphism\[ G\cong K\times H. \]
By step 3, \[ \abs G=\abs K\,\abs H=\abs{K\times H}. \] The injective map \(\alpha\) from step 6 is therefore a bijection between finite groups of equal order, hence an isomorphism.
8. Therefore \(G\) is a direct product of groups of prime order.
By step 5, write \(K\) as a direct product of groups of prime order. The group \(H\) itself has prime order. Step 7 gives \[ G\cong K\times H, \] so adjoining the factor \(H\) gives the required decomposition.
9. Q.E.D.
Steps <1>1–<1>8 complete the induction on \(\abs G\).