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Problem 76 (Homomorphisms Preserve The Identity And Inverses)

Suppose that $f:G\to H$ is a homomorphism, and that $e_G$ and $e_H$ are the identities of $G$ and $H$ respectively.

  1. Prove that $f(e_G)=e_H$.
  2. In addition show that $f(a^{-1})=(f(a))^{-1}$.

Click to see a hint.

You know that $e_Ge_G=e_G$. How does $f(e_Ge_G)=f(e_G)$ help you know that $f(e_G)=e_H$? Remember to use the fact that $f$ is a homomorphism.

To show that $f$ preserves inverses, think about how you can prove that the determinant of an inverse is the inverse of the determinant. If we have two matrices $A$ and $B$ then we know that $|AB|=|A||B|$. If we know that $B=A^{-1}$, then we have $AA^{-1}=I$ and so we have $|A||A^{-1}|=|I|=1$. Since we know that $|A||A^{-1}|=1$, we can just multiply both sides on the left by $|A|^{--1}$ to get $|A^{-1}|=|A|^{-1}$. If you mimic this process with an arbitrary homomorphism, you should find the solution for any homomorphism. Just replace $|A|$ with $f(a)$. What you did in linear algebra is the key to the more generalized notion of a homomorphism.



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