Question

Question: Let $f:G\to H$be an isomorphism of groups. Let $g:H\to G$be the inverse function off as defined in Appendix B. Prove that g is also an isomorphism of groups. [Hint: To show that $g\left(ab\right)=g\left(a\right)g\left(b\right)$, consider the images of the leftand right-hand sides under f and use the facts that f is a homomorphism and $f\circ g$is the identity map.]

Short answer

Answer

It is proved that g is an isomorphism of groups.

Step-by-step solution

Isomorphism of Group 

A one-one and onto homomorphism $f:G\to H$
between two groups G and H is called isomorphism.

Prove that g is an isomorphism of groups

Consider an isomorphism $f:G\to H$from G onto H .

Suppose that $g:H\to G$is the inverse function f .

Claim that the function $g:H\to G$is an isomorphism.

As the function $f:G\to H$is an isomorphism from onto , then the function $g:H\to G$is a bijection from onto .

So, the inverse of exists and is again a bijection. Hence, the function$g:H\to G$is a bijection.

It is sufficient to prove is a homomorphism in order to prove $g:H\to G$the claim for any$x,y\in H$.

As the function is surjective$f:G\to H$, there exist $a,b\in G$such that

$$\begin{gathered} f\left(a\right)=x \\ f\left(b\right)=y \end{gathered}$$

By the definition of inverse of a function,

$$\begin{gathered} g\left(x\right)=a \\ g\left(y\right)=b \end{gathered}$$

As the function $f:G\to H$is a homomorphism,

$$\begin{gathered} f\left(ab\right)=f\left(a\right)f\left(b\right) \\ =xy \end{gathered}$$

Thus,$f^{-1}\left(xy\right)=ab$. This implies $g\left(xy\right)=ab$.

Therefore, $g\left(xy\right)=g\left(x\right)g\left(y\right)forallx,y\in H$

Hence, the function$g:H\to G$ is a homomorphism. This proves the claim.

Therefore, the required result is proved.