Alg7.7

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Prove that a group homomorphism maps the identity to the identity and inverses to inverses.

Since $1=1\cdot 1\,$ the properties of homomorphisms give the relation $\phi(1)=\phi(1\cdot 1) = \phi(1)\phi(1)\,$ .

By applying $\phi(1)^{-1}\,$ to both sides this relation is obtained:

$1=\phi(1)\,$

Next,

$\phi(a^{-1})\phi(a) = \phi(a^{-1}a) = \phi(1) = 1\,$ .

$\phi(a)\phi(a^{-1}) = \phi(a a^{-1}) = \phi(1) = 1\,$ .

Therefore $\phi(a^{-1}) = \phi(a)^{-1}\,$ .

Another proof:

$\phi(1) = \phi(1) \cdot 1 = \phi(1) \cdot (\phi(1) \cdot \phi(1)^{-1}) = (\phi(1) \cdot \phi(1)) \cdot \phi(1)^{-1} = \phi(1 \cdot 1) \cdot \phi(1)^{-1} = \phi(1) \cdot \phi(1)^{-1} = 1$

Next, for $g \in G$

$\phi(g^{-1}) = \phi(g^{-1}) \cdot 1 = \phi(g^{-1}) \cdot ( \phi(g) \cdot \phi(g)^{-1}) = (\phi(g^{-1}) \cdot \phi(g)) \cdot \phi(g)^{-1}$

$= \phi(g \cdot g^{-1}) \cdot \phi(g)^{-1} = \phi(1) \cdot \phi(g)^{-1} = 1 \cdot \phi(g)^{-1} = \phi(g)^{-1}$ .

Abstract Algebra

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