Quotient groups related problem

Click For Summary
SUMMARY

The discussion centers on proving that if \( x^n = e \) for an element \( x \) in a finite group \( G \) with a normal subgroup \( N \) of order \( n \) and \( \text{gcd}(n, [G:N]) = 1 \), then \( x \) must belong to \( N \). The proof involves analyzing the subgroup generated by \( x \) and applying the isomorphism theorem to show that the order of \( M \cap N \) leads to a contradiction if \( x \) is not in \( N \). The participants clarify the conditions and correct misunderstandings regarding group orders and intersections.

PREREQUISITES
  • Understanding of group theory concepts, specifically normal subgroups and quotient groups.
  • Familiarity with the isomorphism theorems in group theory.
  • Knowledge of the properties of finite groups and their orders.
  • Ability to work with gcd (greatest common divisor) in the context of group orders.
NEXT STEPS
  • Study the isomorphism theorems in group theory for a deeper understanding of subgroup relationships.
  • Learn about the structure of finite groups and their normal subgroups.
  • Explore examples of groups with specific orders and their properties, such as the Klein group.
  • Investigate the implications of gcd conditions in group theory proofs.
USEFUL FOR

Mathematics students, particularly those studying abstract algebra and group theory, as well as educators seeking to clarify concepts related to normal subgroups and quotient groups.

zcahana
Messages
7
Reaction score
0

Homework Statement



Let G be a finite group and N\triangleleftG such that |N| = n, and gcd(n,[G:N]) = 1.
Proof that if x^{n} = e then x\inN.

Homework Equations



none.

The Attempt at a Solution



I defined |G| = m and and tried to find an integer which divides both n and m/n.
I went for some X which is not in N, for which X^n = e.
I defined o(X) = q and then q | n.
No luck showing that q | m/n.Any other ideas?Thanks ahead,
Zvi
 
Last edited:
Physics news on Phys.org
This is not true! Take some subgroup H of the Klein group K of order 2. It is normal as K is abelian. K/H also has order 2, and so gcd(H,[K:H]) = 2. But there is certainly an element of order 2 in K that is not in H.

Did you mean to write that the gcd is 1?
 
mrbohn1 said:
Did you mean to write that the gcd is 1?

Yes I did, sorry for the mistake...
*Corrected*
 
Here's a rough sketch:

Suppose x is not in N. Let M=<x>. If xn=e then M has order dividing n. |MN|=mn/|M\capN|, where m is the order of M. MN is a subgroup of G by the second (or third, depending on your numbering!) isomorphism theorem. So mn/|M\capN| divides |G|.

Now, |G| = tn, where t=[G:N] is coprime to n. So m/|M\capN| must divide t. But this is impossible, as m divides n, and t is coprime to n. So x is in N.

As a disclaimer, it's a while since I did any group theory! So you should check all this carefully.
 
Last edited:
Thanks for the response.

By stating that M\capN = <1> did you mean M\capN = e?
If so, it seems to me that it is not necessarily true.
If you meant something else, please elaborate.Zvi.
 
hmmm..yes, sorry about that. I've modified the argument above. Still no guarantee it is right!

Yes, when I say <1> I mean e...itis just to distinguish the identity element from the trivial group.
 
Sorry... I still see a problem with the conclusion that mn divides |G|...
 
That was just an error with my previous editing. Hopefully it is fixed now.

If you spot another problem, try filling in the details yourself...it's the only way to learn! I have done this very much off the cuff, and intended it to be a "suggested method" rather than a full solution to be copied down.
 
Thanks for you help!...
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 26 ·
Replies
26
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K