Where by XHX^H we mean the subgroup ⟨xh=h−1xh|x∈X,h∈H⟩\langle x^h = h^{-1}xh \, | \, x\in X, h \in H \rangle of GG.

What I have so far is:

If xh∈XHx^h \in X^H then clearly xh=h−1xh∈⟨X,H⟩,x^h=h^{-1}xh \in \langle X, H \rangle, so that XH≤⟨X,H⟩X^H \le \langle X, H \rangle. Now to prove normality, I think there are two cases to consider.

If h∈⟨X,H⟩h \in \langle X, H \rangle. Then h−1XHh=(XH)h=XH.h^{-1}X^H h= (X^H)^h = X^H.

If x∈⟨X,H⟩x \in \langle X, H \rangle. Then because x,x−1∈XHx, x^{-1} \in X^H, it follows that x−1XHx=(XH)x=XHx^{-1} X^H x = (X^H)^x=X^H.

Is this correct?

I’m stuck with showing now X⟨X,H⟩≤XHX^{ \langle X, H \rangle} \le X^H. The other side X⟨X,H⟩≥XHX^{ \langle X, H \rangle} \ge X^H is obvious.

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@relep He defined it to be a subgroup.

– Derek Holt

Oct 20 at 19:27

XH⊴⟨X,H⟩⇒X≤NG(XH)⇒(XH)X=X⟨X,H⟩≤XHX^H \unlhd \langle X,H \rangle \Rightarrow X \le N_G(X^H) \Rightarrow (X^H)^X = X^{\langle X,H \rangle} \le X^H.

– Derek Holt

Oct 20 at 19:28

@DerekHolt, I’ve forgotten a bit of group theory. Can you please explain how the implication X≤NG(XH)X \le N_{G}(X^H) follows? Thank you.

– J. Doe

Oct 20 at 19:41

Every subgroup normalizes itself, so it is contained in its own normallizer.

– Derek Holt

Oct 20 at 21:30

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