Group Homomorphism

A function $f:G\to H$ between Groups is a group homomorphism if

• $f(x*y)=f(x)*f(y)$

Properties

• $f(1)=1$

  This follows from a bit of easy algebra

  	$\displaystyle f(1)$	$\displaystyle=f(1)*1$
  		$\displaystyle=f(1)*f(1)*f(1)^{-1}$
  		$\displaystyle=f(1*1)*f(1)^{-1}$
  		$\displaystyle=f(1)*f(1)^{-1}$
  		$\displaystyle=1$

• $f(x^{-1})=f(x)^{-1}$

  Every group is a Cancellative Monoid, so it suffices to show that

  $f(x^{-1})*f(x)=f(x)^{-1}*f(x)$

  This follows from a short calculation

  	$\displaystyle f(x^{-1})*f(x)$	$\displaystyle=f(x^{-1}*x)$
  		$\displaystyle=f(1)$
  		$\displaystyle=1$
  		$\displaystyle=f(x)^{-1}*f(x)$

• If $(H,+)$ is Abelian, then the pointwise sum $f+g:=\lambda x\;f(x)+g(x)$ of two group homomorphisms $f,g:G\to H$ is a group homomorphism:

  	$\displaystyle(f+g)(x+y)$	$\displaystyle=(\lambda x.\;f(x)+g(x))(x+y)$	$\displaystyle(\text{by defn.})$
  		$\displaystyle=f(x+y)+g(x+y)$	$\displaystyle(\text{$\beta$-reduction})$
  		$\displaystyle=f(x)+f(y)+g(x)+g(y)$	$f$ and $g$ are group homomorphisms
  		$\displaystyle=f(x)+g(x)+f(y)+g(y)$	$H$ is abelian
  		$\displaystyle=(\lambda x.\;f(x)+g(x))(x)+(\lambda x.\;f(x)+g(x))(y)$	$\beta$-expansion
  		$\displaystyle=(f+g)(x)+(f+g)(y)$	by defn.

  Note that the sum of two group homomorphisms into a non-abelian group might not be a group homomorphism. This leads to the notion of a Cooperator in a Cartesian Category.