Proving Subgroup Equality in Group Theory

  • Thread starter Thread starter ranoo
  • Start date Start date
AI Thread Summary
To prove that Ha = H if and only if a belongs to H, one must understand that Ha represents the set of all products of elements from H with the element a. If Ha equals H, it implies that every element in H can be expressed as h*a for some h in H, indicating that a must also be in H. For the second part, to show H² = H, it is necessary to demonstrate both H² ⊆ H and H ⊆ H², leveraging the properties of subgroups, particularly closure under multiplication. The closure property ensures that multiplying elements of H will yield elements that remain within H, confirming the equality. Understanding these relationships is crucial for proving subgroup equality in group theory.
ranoo
Messages
9
Reaction score
0

Homework Statement



let H a subgroup of G. Prove that:
1) Ha=H iff a belong to H
2) H2=H

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
ranoo said:

Homework Statement



let H a subgroup of G. Prove that:
1) Ha=H iff a belong to H
2) H2=H
What does Ha mean? What does it mean to multiply a subgroup by itself?
 
I don't know, like this we took in the homework.
 
Ha is the set containing all elements of the form h*a, where h is in H.

What is the definition of a subgroup? It must non-empty, closed under multiplication and closed under inversion. So if Ha = H, then h_{1}a = h_{2}. What does this say about a?

If a is in H on the other hand, and H is closed under multiplication, what does this say about Ha?

Finally, to prove that H^2=H, we need to show that H^{2} \subseteq H and H \subseteq H^{2}. First, naively, which one is definitely contained in which? Next, if H is closed under multiplication, what does this say about the relation between H and H^2?

Good luck!
 
thank you very much
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top