Question

Question: If $f:G\to G$ be a homomorphism of groups, prove that $F=\{a\in G|f\left(a\right)=a\}$is a subgroup of G.

Short answer

It has been proved that $F$is a subgroup of G.

Step-by-step solution

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

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

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

It is surely non-empty since,$f\left(e\right)=e\in F$

Let$a,b\in F$

Since$f$ is a homomorphism

So,

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

This implies$ab\in F$

Hence, the closure propertyis true.

Show that the set is closed under inverses

Now, if $a\in F$

$$\begin{gathered} f\left(a^{-1}\right)=f{\left(a\right)}^{-1} \\ =a^{-1} \end{gathered}$$

So, $a^{-1}\in F$

So, $F$is closed under inverses.

Conclusion

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