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# 392 Actions

 Jun17 accepted Clarification on quotient groups Jun17 asked Clarification on quotient groups Jun17 accepted Congruence relation possible typo? Jun17 asked Congruence relation possible typo? Jun10 accepted Clarification needed on finding last two digits of $9^{9^9}$ Jun10 asked Clarification needed on finding last two digits of $9^{9^9}$ Jun10 accepted Showing a compact metric space has a countable dense subset Jun10 accepted Can a group of order $55$ have exactly $20$ elements of order $11$? (Clarification) Jun9 asked How to determine the parity of a permutation by its cycle decomposition Jun9 revised Can a group of order $55$ have exactly $20$ elements of order $11$? (Clarification) edited body Jun9 asked Can a group of order $55$ have exactly $20$ elements of order $11$? (Clarification) Jun9 comment Can a group of order $55$ have exactly $20$ elements of order $11$? How do you get that there would be 34 elements remaining with order 5? Jun9 comment Showing a compact metric space has a countable dense subset I'm struggling with why $d(x,y)$ has to be less than $1/n$. If the definition of a dense set just requires that another point of $D$ lie in some open neighbourhood of $x \in X$, couldn't it be possible that $y$ is still in $B(x,\epsilon)$ but that the distance between the two points is not less than $1/n$? Jun8 accepted Determining whether or not a group has an element of a specific order Jun8 accepted What can we say about the order of a group given the order of two elements? Jun8 asked Determining whether or not a group has an element of a specific order Jun8 comment What can we say about the order of a group given the order of two elements? To generalize, do I just take all divisors of $n$? If every nontrivial element has the same order, then it must be a $p$-group because the only divisors of $p$ are $1$ and $p$ itself (this means the group $G$ in my problem is certainly not a $p$-group). Is this correct? Jun8 asked Showing a compact metric space has a countable dense subset Jun7 comment What can we say about the order of a group given the order of two elements? could we say it has an element of order 2, 3, and 5? $x^6 = (x^2)^3 = (x^3)^2$, and similarly for $x^{10}$. I know that doesn't say much about $G$, but I'm just trying to figure out what more can be said. Jun7 comment What can we say about the order of a group given the order of two elements? @MartinArgerami Aren't they saying the same thing? Sorry, I was trying to be more formal about it. Thanks for pointing it out