# 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 $H\trianglelefteq G$.

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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