homomorphic image of group

Theorem.  The homomorphic image of a group is a group.  More detailed, if $f$ is a homomorphism from the group  $(G,\ast )$  to the groupoid  $(\mathrm{\Gamma },\star )$,  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

$(\alpha \star \beta )\star \gamma$ $\mathrm{\hspace{0.33em}}=(f(a)\star f(b))\star f(c)$

$\mathrm{\hspace{0.33em}}=f(a\ast b)\star f(c)$

$\mathrm{\hspace{0.33em}}=f((a\ast b)\ast c)$

$\mathrm{\hspace{0.33em}}=f(a\ast (b\ast c))$

$\mathrm{\hspace{0.33em}}=f(a)\star f(b\ast c)$

$\mathrm{\hspace{0.33em}}=f(a)\star (f(b)\star f(c))$

$\mathrm{\hspace{0.33em}}=\alpha \star (\beta \star \gamma ),$

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

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

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

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

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

If  $f(a^{-1})={\alpha }^{\prime }$, then

$$\alpha \star {\alpha }^{\prime }=f(a)\star f(a^{-1})=f(a\ast a^{-1})=f(e)=\epsilon ,$$

$${\alpha }^{\prime }\star \alpha =f(a^{-1})\star f(a)=f(a^{-1}\ast a)=f(e)=\epsilon .$$

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

$$f(n):={⟨4⟩}^n,$$

then $f$ is homomorphism:

$$f(m+n)={⟨4⟩}^{m+n}={⟨4⟩}^m{⟨4⟩}^n=f(m)f(n).$$

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

$$⟨4⟩,⟨16⟩=⟨7⟩,⟨64⟩=⟨1⟩.$$

These apparently form the cyclic group of order 3.

Title	homomorphic image of group
Canonical name	HomomorphicImageOfGroup
Date of creation	2013-03-22 18:56:27
Last modified on	2013-03-22 18:56:27
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	14
Author	pahio (2872)
Entry type	Theorem
Classification	msc 20A05
Classification	msc 08A05
Related topic	GroupHomomorphism
Related topic	CorrespondenceBetweenNormalSubgroupsAndHomomorphicImages