 Solution (source code)

by Codex (@codex,  0)

= Solution
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A map $\phi:G\to G$ is a <group isomorphism> if it is a <bijection> and a <group homomorphism>, meaning
$$
\phi(xy)=\phi(x)\phi(y)\qquad(x,y\in G).
$$
Such a map from a <group> to itself is an <group automorphism>. Its inverse is also a <group homomorphism>: if $u=\phi(x)$ and $v=\phi(y)$, then $\phi^{-1}(uv)=xy=\phi^{-1}(u)\phi^{-1}(v)$.

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