Let \(F\) be a free group and for a fixed integer \(n\), let \(H_n\) be the subgroup generated by the set \(\{ x^n \mid x \in F \}\). Show that \(H_n \trianglelefteq F\).
Solution
1. For all \(g, x \in F\), \(g x^n g^{-1} = (g x g^{-1})^n\).
Conjugation by \(g\) is a homomorphism \(F \to F\), so it sends \(x^n\) to \((g x g^{-1})^n\).
2. For all \(g, x \in F\), \(g x^n g^{-1} \in H_n\).
The element \(g x g^{-1}\) lies in \(F\), so its \(n\)th power \((g x g^{-1})^n\) is one of the generators of \(H_n\). Step 1 identifies this power with \(g x^n g^{-1}\).
3. For every \(g \in F\), \(g H_n g^{-1} \subseteq H_n\).
Conjugation by \(g\) is a homomorphism, so \(g H_n g^{-1}\) is the subgroup generated by the conjugates \(g x^n g^{-1}\) of the generators of \(H_n\). By step 2 these conjugates lie in \(H_n\), hence so does the subgroup they generate.
4. Q.E.D.
Step 3 is the definition of \(H_n \trianglelefteq F\).