A detail in a proof about isomorphism classes of groups of order 21

In summary, the conversation discusses the existence of two isomorphism classes of groups of order 21 and a proof of this statement. The conversation then moves on to discussing the Third Sylow Theorem and the question of why a certain subgroup must be normal. The expert summarizer suggests using the definition of normal and the fact that there is only one subgroup of order 7 to show that this subgroup is indeed normal.
  • #1
Whovian
652
3

Homework Statement



While reading through my textbook on abstract algebra while studying for a test, I ran across the following statement:

There are two isomorphism classes of groups of order 21: the class of ##C_{21}##, and the class of a group ##G## generated by two elements ##x## and ##y## that satisfy the relations ##x^7=1##, ##y^3=1##, ##yx=x^2y##

along with a proof of this statement. The proof looks fairly reasonable, only there's a bit I don't quite understand:

"The Third Sylow Theorem shows us that the Sylow 7-subgroup ##K## [existence and uniqueness guaranteed by the Third Sylow Theorem] must be normal ..."

Specifically, why must ##K\trianglelefteq G##?

Homework Equations



Third Sylow Theorem: If ##G## is a group of order ##n<\aleph_0##, where ##n=p^e\cdot m## for some ##e>0## and ##\lnot\left(p|m\right)##, for some prime ##p##, and ##s## is the number of Sylow ##p##-subgroups of ##G##, then ##s|m## and ##s\equiv1\pmod p##.

The Attempt at a Solution



Fairly nonexistent; I can't think of any reason this subgroup must be normal.

(If this post would be more appropriate in the linear & abstract algebra section, feel free to move it there.)
 
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  • #2
Whovian said:

Homework Statement



While reading through my textbook on abstract algebra while studying for a test, I ran across the following statement:

There are two isomorphism classes of groups of order 21: the class of ##C_{21}##, and the class of a group ##G## generated by two elements ##x## and ##y## that satisfy the relations ##x^7=1##, ##y^3=1##, ##yx=x^2y##

along with a proof of this statement. The proof looks fairly reasonable, only there's a bit I don't quite understand:

"The Third Sylow Theorem shows us that the Sylow 7-subgroup ##K## [existence and uniqueness guaranteed by the Third Sylow Theorem] must be normal ..."

Specifically, why must ##K\trianglelefteq G##?

Homework Equations



Third Sylow Theorem: If ##G## is a group of order ##n<\aleph_0##, where ##n=p^e\cdot m## for some ##e>0## and ##\lnot\left(p|m\right)##, for some prime ##p##, and ##s## is the number of Sylow ##p##-subgroups of ##G##, then ##s|m## and ##s\equiv1\pmod p##.

The Attempt at a Solution



Fairly nonexistent; I can't think of any reason this subgroup must be normal.

(If this post would be more appropriate in the linear & abstract algebra section, feel free to move it there.)

Use what they are saying about ##s## to show there must be only one 7-subgroup. So?
 
  • #3
There must be a 7-subgroup; I just don't see why it must be a normal subgroup.
 
  • #4
Whovian said:
There must be a 7-subgroup; I just don't see why it must be a normal subgroup.

Use the definition of normal. There is only ONE 7-subgroup.
 
  • #5
I'm used to normal being defined as "closed under conjugation by elements of G." Is there a different or equivalent definition I'm missing or something else obvious which makes clear that only one 7-subgroup existing means said subgroup is normal? (Knowing me, most likely.)
 
  • #6
If you conjugate a subgroup, the result is another subgroup of the same size. So if there's only one subgroup of order 7...
 

FAQ: A detail in a proof about isomorphism classes of groups of order 21

1. What is a proof about isomorphism classes of groups of order 21?

A proof about isomorphism classes of groups of order 21 is a mathematical argument that shows how all groups of order 21 are related to each other through isomorphism. This means that these groups have the same underlying structure, even though they may appear different on the surface.

2. Why is it important to study groups of order 21?

Studying groups of order 21 allows us to gain a deeper understanding of group theory and its applications in various fields such as mathematics, physics, and computer science. It also helps us in solving more complex problems involving groups of higher orders.

3. How does the proof for isomorphism classes of groups of order 21 work?

The proof for isomorphism classes of groups of order 21 involves breaking down the groups into smaller subgroups and analyzing their properties. By using various theorems and techniques, it is possible to show that these groups are isomorphic to each other.

4. What are some examples of groups of order 21?

Some examples of groups of order 21 include the cyclic group of order 21, the dihedral group of order 21, and the alternating group of degree 7. These groups have different structures but are all isomorphic to each other.

5. Can the proof for isomorphism classes of groups of order 21 be applied to other orders?

Yes, the proof for isomorphism classes of groups of order 21 can be generalized to other orders as well. However, the techniques and theorems used may differ depending on the specific order of the groups being studied.

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