Abstract Algebra: Groups and Subgroups

AI Thread Summary
The discussion focuses on proving that the set H, consisting of elements from S that commute with every element in S, is closed under the binary operation *. Participants express confusion about how to approach the problem, particularly regarding the associative property and the lack of need for isomorphism. The key to the solution lies in demonstrating that for any elements a and b in H, the product a * b also belongs to H, using the definition of H and the associative property. Understanding these concepts is crucial for successfully tackling similar problems on the exam. Clarity on these foundational ideas will enhance comprehension of group theory in abstract algebra.
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Homework Statement


The problem says: Suppose that * is an associative binary operation on a set S.
Let H= {a ε S l a * x = x * a for all x ε s}. Show that H is closed under *. ( We think of H as consisting of all elements of S that commute with every element in S)

My teacher is horrible so I am pretty lost in the class. I am aware of what the associative property is, but I'm not sure how to go about solving this question when it comes to the binary operation. This is going to be on my exam so I need to know how to solve it.
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Homework Equations


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The Attempt at a Solution


I know that with associative and groups you would try to prove its isomorphism but I'm not sure where to begin with this one
 
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No isomorphism required...

You're just trying to show that, for any a,b\inH, a*b\inH. You should be able to do this with your definition of H plus associativity.
 
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
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