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Exercise (The set of simple shift permutations on 26 letters is isomorphic to $\mathbb{Z}_{26}$.)
Prove that the function $f:H_{26}\to \mathbb{Z}_{26}$ defined by $f(\phi_j)=j$ is a group isomorphism.
Click to see a solution.
We have to show two things. We must show that the function is a bijection, and that the function preserves the group operations.
- This function is a bijection because it has an inverse, namely $f^{-1}(k)=\phi_k$.
- We now have to show that the functin perserves the group operation. To show this we compute
$$f(\phi_j\circ \phi_k) = f(\phi_{j+k})= f(\phi_{(j+k)\pmod {26}}) = (j+k)\pmod {26} = (f(\phi_j)+f(\phi_k))\pmod {26}.$$ This shows that $f(\phi_j\circ \phi_k) = (f(\phi_j)+f(\phi_k))\pmod {26}$, which means the the function preserves the group operation.