Question

Question: Let G, H and K be groups. If$\mathit{G}\mathbf{\cong }\mathit{H}$and $\mathit{H}\mathbf{\cong }\mathit{K}$, then prove that $\mathit{G}\mathbf{\cong }\mathit{K}$. [Hint: If and are isomorphisms, prove that the composite function is also an isomorphism.]

Short answer

It has been proved that$G\cong K$.

Step-by-step solution

Step-By-Step SolutionStep 1: Define a composite map

Let$f:G\to H$ and$g:H\to K$are isomorphisms.

Define a composite function$g\circ f:G\underset{}{\to }K$

Show that is injective

Consider ,$g\left(f\left(x\right)\right)=g\left(f\left(y\right)\right)$

Since gis injective, we get $f\left(x\right)=f\left(y\right)$

Since f is injective, we get $x=y$

Hence $g\circ f$is injective.

Show that is surjective

Let $k\in K$

Since g is surjective, there exists some $h\in H$with $g\left(h\right)=k$.

Since f is surjective, there exists some $x\in G$with $f\left(x\right)=h$.

so ,

$$\begin{gathered} \left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left(h\right) \\ =k \end{gathered}$$

Hence, $g\circ f$is surjective.

Show that is homomorphism

Since both f and g are homomorphisms,

Consider

$$\begin{gathered} \left(g\circ f\right)\left(xy\right)=g\left(f\left(xy\right)\right) \\ =g\left(f\left(x\right)f\left(y\right)\right) \\ =g\left(f\left(x\right)\right)g\left(f\left(y\right)\right) \\ =g\circ f\left(x\right)g\circ f\left(y\right) \end{gathered}$$

Hence, $g\circ f$is a homomorphism.

Conclusion

Hence, $g\circ f$is an isomorphism.

thus,$G\cong K$