#### quasar987

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So how would the proof go w/o this silly notation? It seems we still have to show the same thing, namely, that

[tex]h(a\cdot b) = h(a) \circ h(b)[/tex]

since the definition of homomorphism is independant of the notation we use. But now the problem is genuine, since

[tex]h(a) \circ h(b) = h_a(gb)[/tex]