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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.

Solution

Consider $f( e_G )$. Since $f$ is a homomorphism between $G$ and $H$ we know that $f$ preserves the group structure, therefore we compute $$ \begin{align} f( e_G ) &= f( e_G \cdot e_G ) \\ &= f( e_G ) \times f( e_G )\; . \end{align} $$ Seeing that $f( e_G ) = f( e_G ) \times f( e_G )$ this statement is clearly equivalent to $e_H \times f( e_G ) = f( e_G ) \times f( e_G )$, and since $H$ is a group we may apply cancellation laws to get $$ e_H = f( e_G ). $$ Q.E.D.

Let $a,b \in G$ such that $b = a^{-1}$, the inverse of $a$ in $G$. We then compute $$ \begin{align} e_H &= f( e_G ) \\ &= f( a \cdot b ) \\ &= f( a ) \times f( b ) \\ &= f( a ) \times f( a^{-1} )\; . \end{align} $$ Seeing that $e_H = f( a ) \times f( a^{-1} )$ and since $H$ is a group then we know that the inverse of $f( a )$ is in $H$, namely $( f( a ) )^{-1}$. Therefore, left multiplying by $( f( a ) )^{-1}$ yields $$ \begin{align} ( f(a) )^{-1} \times e_H = ( f(a) )^{-1} \times f(a) \times f(a^{-1}) &\Longleftrightarrow ( f(a) )^{-1} = e_H \times f(a^{-1}) \\ &\Longleftrightarrow ( f(a) )^{-1} = f( a^{-1} )\; . \end{align} $$ Q.E.D.

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