Putnam 1997 A4 Session A Let GGG be a group with identity eee and ϕ:G→G\phi:G\rightarrow Gϕ:G→G a function such that ϕ(g1)ϕ(g2)ϕ(g3)=ϕ(h1)ϕ(h2)ϕ(h3)\phi(g_1)\phi(g_2)\phi(g_3)=\phi(h_1)\phi(h_2)\phi(h_3)ϕ(g1)ϕ(g2)ϕ(g3)=ϕ(h1)ϕ(h2)ϕ(h3) whenever g1g2g3=e=h1h2h3g_1g_2g_3=e=h_1h_2h_3g1g2g3=e=h1h2h3. Prove that there exists an element a∈Ga\in Ga∈G such that ψ(x)=aϕ(x)\psi(x)=a\phi(x)ψ(x)=aϕ(x) is a homomorphism (i.e. ψ(xy)=ψ(x)ψ(y)\psi(xy)=\psi(x)\psi(y)ψ(xy)=ψ(x)ψ(y) for all x,y∈Gx,y\in Gx,y∈G). Curated text Kedlaya Putnam archive. Quoted with source. Solutions Human solution Solution 1 by Manjul Bhargava, Kiran Kedlaya and Lenny Ng The terms of the source let olymmath link to this solution but not copy it. Read the solution (Kedlaya Putnam archive). PreviousA3 NextA5 Locator putnam/1997/a4 · Tag 83N4A3MFTT · Edit this page on GitHub · Keys: j next, k previous
Human solution Solution 1 by Manjul Bhargava, Kiran Kedlaya and Lenny Ng The terms of the source let olymmath link to this solution but not copy it. Read the solution (Kedlaya Putnam archive).