Discover the Cyclic Subgroups of U(30): Complete Guide

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The discussion centers on identifying the cyclic subgroups of U(30). The user initially lists seven subgroups, including <17> and <23>, but the answer key only includes five subgroups, omitting <17> and <23>. The user seeks clarification on their error, with a suggestion that duplicates may be present in their list. The conversation highlights the importance of verifying subgroup definitions and ensuring no repetitions occur. Understanding the structure of U(30) is crucial for accurately determining its cyclic subgroups.
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Please help me.

The question is: List the cyclic Subgroups of U(30)

My answer {<1>,<7>,<13>,<17>,<19>,<23>,<29>} however the answer key tells me {<1>,<7>,<13>,<19>,<29>} (leaving out <17>,<23>)

Where am I erroneous?
 
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where is erroneous?

The question is: List the cyclic Subgroups of U(30)

My answer {<1>,<7>,<13>,<17>,<19>,<23>,<29>} however the answer key tells me {<1>,<7>,<13>,<19>,<29>} (leaving out <17>,<23>)

Where am I erroneous?
 
You probably have duplicates.
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

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
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Thread 'Greatest possible value of a constant in polynomial'

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
View full post »

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