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 (aa^{{-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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