# Showing that subgroup of unique order implies normality

## 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##.

## 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##.

fresh_42
Mentor

## 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##.

## 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}|##.
Right, but the reason for your subgroup argument is the same as the direct line, which was asked here:
## (ghg^{-1})\cdot (gkg^{-1})^{-1}=ghg^{-1}gk^{-1}g^{-1}=ghk^{-1}g^{-1} ## which is no big deal to write instead.
(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##.
This is quite obvious and no proof is needed. If ##A=B## you can always substitute these letters in any statement.
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##.
This is not what (*) said, at least I don't see it. But anyway. There is again no proof needed. ##H=gHg^{-1}## is the definition of a normal subgroup. If at all, you should have written ##a(gHg^{-1})a = (ag)H(ag)^{-1}## and ##\{\,ag\, : \,g\in G\,\}=\{\,g\, : \,g\in G\,\}## for all ##a\in G##.

The essential part is ##gHg^{-1}=H## by the uniqueness of subgroups of order ##|H|##, a condition which I like to note is rarely given. Usually there are a lot of isomorphic copies of a subgroup also subgroups.

Your proof is correct, even though a bit too detailed at unnecessary locations, and a bit too general at locations, where a small calculation would have done it.

Mr Davis 97
Right, but the reason for your subgroup argument is the same as the direct line, which was asked here:
## (ghg^{-1})\cdot (gkg^{-1})^{-1}=ghg^{-1}gk^{-1}g^{-1}=ghk^{-1}g^{-1} ## which is no big deal to write instead.

This is quite obvious and no proof is needed. If ##A=B## you can always substitute these letters in any statement.

This is not what (*) said, at least I don't see it. But anyway. There is again no proof needed. ##H=gHg^{-1}## is the definition of a normal subgroup. If at all, you should have written ##a(gHg^{-1})a = (ag)H(ag)^{-1}## and ##\{\,ag\, : \,g\in G\,\}=\{\,g\, : \,g\in G\,\}## for all ##a\in G##.

The essential part is ##gHg^{-1}=H## by the uniqueness of subgroups of order ##|H|##, a condition which I like to note is rarely given. Usually there are a lot of isomorphic copies of a subgroup also subgroups.

Your proof is correct, even though a bit too detailed at unnecessary locations, and a bit too general at locations, where a small calculation would have done it.
In the case that I just used ##(ghg^{-1})\cdot (gkg^{-1})^{-1}=ghg^{-1}gk^{-1}g^{-1}=ghk^{-1}g^{-1}## to show that ##gHg^{-1}## is a subgroup, how could I show easily that it has the same order as ##H## without again having to reference the inner automorphism ##c_g##?

fresh_42
Mentor
In the case that I just used ##(ghg^{-1})\cdot (gkg^{-1})^{-1}=ghg^{-1}gk^{-1}g^{-1}=ghk^{-1}g^{-1}## to show that ##gHg^{-1}## is a subgroup, how could I show easily that it has the same order as ##H## without again having to reference the inner automorphism ##c_g##?
No, this part was alright. ##h\longmapsto ghg^{-1}## is a bijection - that was it, and you wrote it. I just wanted to say that the line for the subgroup property is basically the same line that gives you the homomorphism property you used. So instead of quoting a Lemma, this line would have done the same job, without having to verify that sugroups go to subgroups under homomorphisms. But, sure, your argument was correct!

Mr Davis 97