Question

If G and Hare groups, prove that the function$f:G\times H\to G$ given by $f(a,b)=a$is a surjective homomorphism.

Short answer

The function $f:G\times H\to G$defined by$f(a,b)=a$is a surjective homomorphism.

Step-by-step solution

Definition of Isomorphism

Let G and H be the groups with the group operation denoted by $*$. Group G is isomorphic to group H if there is a function $f:G\to H$such that,

(i) f is injective

(ii) f is surjective

(iii) $f\left(ab\right)=f\left(a\right)f\left(b\right)$for all$a,b\in G$

f is surjective

If $a\in G$, it is obvious that $f(a,e_H)=a$.

Hence, f is surjective.

f is a homomorphism

Let $(a,b),(c,d)\in G\times H$then,

$$\begin{gathered} f\left(\right(a,b)*(c,d\left)\right)=f\left(\right(a*c),(b*d\left)\right) \\ =(a*c) \\ =f(a,b)*f(c,d) \end{gathered}$$

Thus, for all $(a,b),(c,d)\in G\times H,f\left(\right(a,b)*(c,d\left)\right)=f(a,b)*f(c,d)$, and hence, f is a homomorphism.

Therefore, the function f:$G\times H\to G$ defined by $f(a,b)=a$is a surjective homomorphism.