Question

Show that identities are equal in a group homomorphism

Answer 1

Answer:

Step-by-step explanation:

Theorem

Let \left({G, \circ}\right) and \left({H, *}\right) be groups.

Let \phi: \left({G, \circ}\right) \to \left({H, *}\right) be a group homomorphism.

Let:

e_G be the identity of G

e_H be the identity of H.

Then:

\phi \left({e_G}\right) = e_H

Proof 1

\displaystyle \map \phi {e_G} =    \displaystyle \map \phi {e_G \circ e_G}   \quad Definition of Identity Element \quad

\displaystyle =    \displaystyle \map \phi {e_G} * \map \phi {e_G}   \quad Definition of Morphism Property \quad

That is:

\displaystyle \map \phi {e_G} * e_H =    \displaystyle \map \phi {e_G} * \map \phi {e_G}   \quad Definition of Identity Element \quad

\displaystyle \leadsto \ \  \displaystyle e_H =    \displaystyle \map \phi {e_G}   \quad Cancellation Laws \quad

\blacksquare

Proof 2

A direct application of Homomorphism to Group Preserves Identity.

\blacksquare

Proof 3

From Group Homomorphism of Product with Inverse, we have:

\forall x, y \in G: \phi \left({x \circ y^{-1}}\right) = \phi \left({x}\right) * \left({\phi \left({y}\right)}\right)^{-1}

Putting x = y we have:

\displaystyle \phi \left({e_G}\right) =    \displaystyle \phi \left({x \circ x^{-1} }\right)   \quad \quad

\displaystyle =    \displaystyle \phi \left({x}\right) * \left({\phi \left({x}\right)}\right)^{-1}   \quad \quad

\displaystyle =    \displaystyle e_H   \quad \quad

\blacksquare