- #1

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## Homework Statement

Let ##H## be a subgroup of ##G## and fix some element ##g\in G##.

Prove that ##gHg^{-1}=\{ghg^{-1} \mid h\in H\}## is a subgroup of ##G## of the same order as ##H##.

Deduce that if ##H## is the unique subgroup of ##G## of order ##|H|## then ##H\trianglelefteq G##.

## Homework Equations

## The Attempt at a Solution

(a) Consider the inner automorphism of ##G##, ##c_g(x)=gxg^{-1}##. In particular, ##c_g## is a homomorphism. We know that the homomorphic image of a subgroup is also a subgroup, so we have that ##c_g(H) \le G \implies gHg^{-1} \le G##. Also, in particular, ##c_g## is a bijection from ##G## to ##G##. So by set theory, if ##A\subseteq G##, then ##|A|=|c_g(A)|=|gAg^{-1}|##. Since in particular ##H\subseteq G##, ##|H|=|gHg^{-1}|##.

(b) Lemma *: If ##A,B\subseteq G##, ##A=B##, and ##g\in G##, then ##Ag=Bg##. Suppose that ##x\in Ag##. Then ##x=ag## for some ##a\in A##. But ##a## is also in ##B##, since ##A=B##. So ##x=ag\in Bg##. The reverse containment is nearly identical, just swap the ##a## and the ##b##. So ##Ag=Bg##.

Now, suppose that ##H## is the unique subgroup of G of order ##|H|##. By part (a) we have that ##|gHg^{-1}| = |H|## and that ##gHg^{-1}\le G##, so by the uniqueness of the subgroup ##H## having order ##|H|##, we must have that ##H=gHg^{-1}##. Using lemma *, we see that ##gH=Hg##. But our fixing of ##g## was arbitrary, so ##gH=Hg## is true for all ##g\in G##. This means that ##H\trianglelefteq G##.