Question

Question: Let $f:G\to H$be a homomorphism of groups and let Kbe a subgroup of H. Prove that the set $\{a\in G|f\left(a\right)\in K\}$is a subgroup of G.

Short answer

It has been proved that the set $\{a\in G|f\left(a\right)\in K\}$is a subgroup of G.

Step-by-step solution

Step-By-Step SolutionStep 1: Show that the set is closed under operation

Given that $f:G\to H$is a homomorphism of groups.

K is a subgroup of H.

Consider$A=\{a\in G|f\left(a\right)\in K\}$

Let $a,b\in A$, then$f\left(a\right),f\left(b\right)\in K$

So,$f\left(ab\right)=f\left(a\right)f\left(b\right)$

$f\left(a\right)f\left(b\right)\in K$implies$ab\in A$

Hence, the closure property is true.

Show that the set is closed under inverses

Now,$f\left(a^{-1}\right)=f{\left(a\right)}^{-1}\in K$

so,$a^{-1}\in A$

So, Ais closed under inverses.

Conclusion

Since, Ais non-empty and closed under operation and inverses, therefore, it is a subgroup.