Mapping of power is power of mapping

Dependencies:

  1. Homomorphism on groups

Let $\phi$ be a homomorphism from $G$ to $H$. Let $g \in G$. Then $\phi(g^k) = \phi(g)^k$.

Proof

For $k > 0$, \[ \phi(g^k) = \phi(gg\ldots g) = \phi(g)\phi(g)\ldots\phi(g) = \phi(g)^k \]

For $k = 0$, \[ e_Ge_G = e_G \Rightarrow \phi(e_G)\phi(e_G) = \phi(e_G) \Rightarrow \phi(e_G) = e_H \Rightarrow \phi(g^k) = \phi(g)^k \]

For $k < 0$, \[ e_H = \phi(e_G) = \phi(gg^{-1}) = \phi(g)\phi(g^{-1}) \implies \phi(g^{-1}) = \phi(g)^{-1} \] \[ \phi(g^k) = \phi((g^{-1})^{-k}) = \phi(g^{-1})^{-k} = (\phi(g)^{-1})^{-k} = \phi(g)^k \]

In all 3 cases, $\phi(g^k) = \phi(g)^k$.

Dependency for:

  1. First isomorphism theorem
  2. Order of elements is invariant under isomorphism
  3. Cyclicness is invariant under isomorphism
  4. Homomorphic mapping of subgroup of domain is subgroup of codomain

Info:

Transitive dependencies:

  1. Group
  2. Homomorphism on groups