group homomorphism

**Greengoblin** · February 21st 2009, 11:57 AM

If $a:G\to H$ is a group homomorphism between G and H, then we have the following definition:

for all $g,k\in G$ we have $a(g*k)=a(g)*a(k)$

According to wikipedia, this means that $a(1_G)=1_H$ , where $1_G,1_H$ , are the identity elements of G and H respectively, and also that $a(g^{-1})=a(g)^{-1}$ for all $g\in G$ .

Can someone show me how these conditions are true?

Thanks

**clic-clac** · February 21st 2009, 12:45 PM

Hi

One way is to write $a(1_G)=a(1_G1_G)=a(1_G)a(1_G)$ thus $a(1_G)=1_H.$

Now that we know that, $a(g^{-1})a(g)=a(g^{-1}g)=a(1_G)=1_H$ implies that $a(g^{-1})=(a(g))^{-1}.$

**Greengoblin** · February 21st 2009, 01:29 PM

Thanks, but I don't understand how your first argument shows that $a(1_G)=1_H$

**clic-clac** · February 21st 2009, 02:08 PM

Ok, that's because a group law is strong enough: let $A$ be a group, $e$ its identity element and $a\in A,$ then $a^2=a\Rightarrow a=e$
Indeed, $a^2=a\Rightarrow a=a^{-1}a^2=a^{-1}a=e$

That may be a particular case of: $\forall a,b\in A,\ \exists !x\ \text{s.t.}\ ax=b$

Indeed if $b=a,$ since $e$ is a solution, by unicity, $aa=a\Rightarrow a=e$

**Greengoblin** · February 22nd 2009, 09:54 AM

Thanks, I got it now.

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