# Group Homomorphism Mapping

• April 12th 2008, 06:16 AM
moolimanj
Group Homomorphism Mapping
Hi all, need to get this done by tonight. Any help would be muh appreciated
• April 12th 2008, 07:36 AM
moolimanj
Hi, has this got anything to do with proving the inverse, identity and closure axioms? If so, can someone start me off please?
• April 12th 2008, 07:43 PM
ThePerfectHacker
Quote:

Originally Posted by moolimanj
Hi, has this got anything to do with proving the inverse, identity and closure axioms? If so, can someone start me off please?

Show what you did. Show how you applied the definitions of homomorphism here.
• April 13th 2008, 03:11 AM
moolimanj
Can you start me off please - i really dont know where to begin
• April 13th 2008, 08:37 AM
ThePerfectHacker
Your mapping does not make so much sense. What do $\phi: G\times G\mapsto H$ mean? It can only make sense if $H = G\times G$. Because $\phi(g_1,g_2) = (g_1,g_1)$ and this needs to be an element of $H$, for that to happen we need $H=G\times G$. Thus, I will assume that.

The first step is to show this is a group homomorphism. $\phi ((g_1,g_2)(g_1',g_2')) = \phi(g_1g_1',g_2g_2') = (g_1g_1',g_2g_2') = (g_1,g_2)(g_1',g_2') = \phi(g_1,g_2)\phi(g_1',g_2')$.