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Jun
17
asked Open sets in $\Bbb{R}^2$ and $\Bbb{R}^3$
Jun
17
comment Topologist's sine curve is connected
Doesn't disconnected mean that there are two disjoint open sets $A$ and $B$ such that the entire set $S$ is equal to $A\cup B$? But when you take your definitions of $A$ and $B$, the union of them is more than what is needed, no?
Jun
17
accepted Clarification on quotient groups
Jun
17
asked Clarification on quotient groups
Jun
17
accepted Congruence relation possible typo?
Jun
17
asked Congruence relation possible typo?
Jun
10
accepted Clarification needed on finding last two digits of $9^{9^9}$
Jun
10
asked Clarification needed on finding last two digits of $9^{9^9}$
Jun
10
accepted Showing a compact metric space has a countable dense subset
Jun
10
accepted Can a group of order $55$ have exactly $20$ elements of order $11$? (Clarification)
Jun
9
asked How to determine the parity of a permutation by its cycle decomposition
Jun
9
revised Can a group of order $55$ have exactly $20$ elements of order $11$? (Clarification)
edited body
Jun
9
asked Can a group of order $55$ have exactly $20$ elements of order $11$? (Clarification)
Jun
9
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?
Jun
9
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 $1/n$?
Jun
8
accepted Determining whether or not a group has an element of a specific order
Jun
8
accepted What can we say about the order of a group given the order of two elements?
Jun
8
asked Determining whether or not a group has an element of a specific order
Jun
8
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?
Jun
8
asked Showing a compact metric space has a countable dense subset