# Showing that subgroup of unique order implies normality

• Mr Davis 97
In summary: But that's not really a critique, sometimes people like to be redundant.You could have used this argument to show that ##|gHg^{-1}|=|H|## instead of the bijection, but I don't see how this is easier. You would have to show that ##(gHg^{-1})^2=H## and so on, and so on. It's a bit more work. But we are talking about lines of length 1 here, no real difference.##(ghg^{-1})\cdot (gkg^{-1})^{-1} = ghg^{-1}gk^{-1}g^{-1}## is just
Mr Davis 97

## 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##.

Mr Davis 97 said:

## 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
fresh_42 said:
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##?

Mr Davis 97 said:
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

## 1. What is a subgroup of unique order?

A subgroup of unique order is a subset of a larger group that contains a specific number of elements, and no other subgroup of the same group has the same number of elements. For example, a subgroup of unique order 3 would have 3 elements and no other subgroup in the same group would have 3 elements.

## 2. How does a subgroup of unique order imply normality?

A subgroup of unique order implies normality because it satisfies the definition of a normal subgroup. A normal subgroup is a subgroup that is invariant under conjugation by any element in the larger group. In other words, if the elements of the subgroup are rearranged by any element in the larger group, the resulting subgroup is still the same. This is true for subgroups of unique order because they have a specific number of elements and cannot be rearranged to form a different subgroup.

## 3. Can a subgroup of unique order exist in any group?

Yes, a subgroup of unique order can exist in any group. However, not all groups will have subgroups of unique order. For example, a group with only prime order elements will not have any subgroups of unique order.

## 4. How can you prove that a subgroup of unique order is normal?

To prove that a subgroup of unique order is normal, you can use the definition of a normal subgroup and show that the subgroup is invariant under conjugation by any element in the larger group. This can be done by showing that for any element in the subgroup and any element in the larger group, their product will also be in the subgroup.

## 5. What is the significance of proving that a subgroup of unique order is normal?

Proving that a subgroup of unique order is normal is significant because it is a useful tool in group theory. It allows us to identify and analyze the normal subgroups of a group, which can then be used to simplify and understand the structure of the larger group. This can also help in proving other theorems and properties about the group.

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