• Support PF! Buy your school textbooks, materials and every day products Here!

Showing that subgroup of unique order implies normality

  • #1
1,456
44

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##.
 

Answers and Replies

  • #2
12,654
9,181

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}|##.
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.
 
  • #3
1,456
44
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##?
 
  • #4
12,654
9,181
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!
 

Related Threads on Showing that subgroup of unique order implies normality

Replies
4
Views
3K
Replies
4
Views
844
Replies
0
Views
5K
Replies
3
Views
9K
Replies
3
Views
4K
Replies
4
Views
558
Replies
1
Views
396
Replies
0
Views
882
Top