# homomorphic image of group

The homomorphic image of a group is a group.  More detailed, if $f$ is a homomorphism from the group  $(G,\,\ast)$  to the groupoid$(\Gamma,\,\star)$,  then the groupoid  $(f(G),\,\star)$  also is a group.  Especially, the isomorphic image of a group is a group.

Proof.  Let $\alpha,\,\beta,\,\gamma$ be arbitrary elements of the image $f(G)$ and $a,\,b,\,c$ some elements of $G$ such that  $f(a)=\alpha,\,f(b)=\beta,\,f(c)=\gamma$.  Then

 $\alpha\star\beta\;=\;f(a)\star f(b)\;=\;f(a\ast b)\;\in\;f(G),$

whence $f(G)$ is closed under$\star$”, and we, in fact, can speak of a groupoid  $(f(G),\,\star)$.

Secondly, we can calculate

 $\displaystyle(\alpha\star\beta)\star\gamma$ $\displaystyle\;=\;(f(a)\star f(b))\star f(c)$ $\displaystyle\;=\;f(a\ast b)\star f(c)$ $\displaystyle\;=\;f((a\ast b)\ast c)$ $\displaystyle\;=\;f(a\ast(b\ast c))$ $\displaystyle\;=\;f(a)\star f(b\ast c)$ $\displaystyle\;=\;f(a)\star(f(b)\star f(c))$ $\displaystyle\;=\;\alpha\star(\beta\star\gamma),$

whence the associativity is in in the groupoid $(f(G),\,\star)$.

Let $e$ be the identity element of  $(G,\,\ast)$  and  $f(e)=\varepsilon$.  Then

 $\varepsilon\star\alpha\;=\;f(e)\star f(a)\;=\;f(e\ast a)\;=\;f(a)\;=\;\alpha,$
 $\alpha\star\varepsilon\;=\;f(a)\star f(e)\;=\;f(a\ast e)\;=\;f(a)\;=\;\alpha,$

and therefore $\varepsilon$ is an identity element in $f(G)$.

If  $f(a^{-1})=\alpha^{\prime}$, then

 $\alpha\star\alpha^{\prime}\;=\;f(a)\star f(a^{-1})\;=\;f(a\ast a^{-1})\;=\;f(e% )\;=\;\varepsilon,$
 $\alpha^{\prime}\star\alpha\;=\;f(a^{-1})\star f(a)\;=\;f(a^{-1}\ast a)\;=\;f(e% )\;=\;\varepsilon.$

Thus any element $\alpha$ of $f(G)$ has in $f(G)$ an inverse.

Accordingly,  $(f(G),\,\star)$  is a group.

Remark 1.  If  $(G,\,\ast)$  is Abelian, the same is true for  $(f(G),\,\star)$.

Remark 2.  Analogically, one may prove that the homomorphic image of a ring is a ring.

Example.  If we define the mapping $f$ from the group  $(\mathbb{Z},\,+)$  to the groupoid  $(\mathbb{Z}_{9},\,\cdot)$  by

 $f(n)\;:=\;\langle 4\rangle^{n},$

then $f$ is homomorphism:

 $f(m\!+\!n)\;=\;\langle 4\rangle^{m+n}\;=\;\langle 4\rangle^{m}\langle 4\rangle% ^{n}\;=\;f(m)f(n).$

The image $f(\mathbb{Z})$ consists of powers of the residue class (http://planetmath.org/Congruences) $\langle 4\rangle$, which are

 $\langle 4\rangle,\;\;\langle 16\rangle=\langle 7\rangle,\;\;\langle 64\rangle=% \langle 1\rangle.$

These apparently form the cyclic group of order 3.

Title homomorphic image of group HomomorphicImageOfGroup 2013-03-22 18:56:27 2013-03-22 18:56:27 pahio (2872) pahio (2872) 14 pahio (2872) Theorem msc 20A05 msc 08A05 GroupHomomorphism CorrespondenceBetweenNormalSubgroupsAndHomomorphicImages