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  • 0 posts edited
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Nov
25
asked The kernel of a group action defined by left multiplication on left cosets of $H\leq G$
Nov
24
comment What can be said about $G$ non-abelian if $\mathbb{Z}_2\rightarrow G\rightarrow\mathbb{Z}_2^3$ and $Z(G)\cong\mathbb{Z}_2$?
It's true if the center is isomorphic to $\mathbb{Z}_2$. Yes I know there are 9 non-abelian groups of order 16, but I was able to rule out 7 of them.
Nov
24
comment What can be said about $G$ non-abelian if $\mathbb{Z}_2\rightarrow G\rightarrow\mathbb{Z}_2^3$ and $Z(G)\cong\mathbb{Z}_2$?
ok cool thanks =).
Nov
24
comment What can be said about $G$ non-abelian if $\mathbb{Z}_2\rightarrow G\rightarrow\mathbb{Z}_2^3$ and $Z(G)\cong\mathbb{Z}_2$?
Ok thanks Alexander, any thoughts on how to prove it besides just trying every group of order 16?
Nov
23
revised What can be said about $G$ non-abelian if $\mathbb{Z}_2\rightarrow G\rightarrow\mathbb{Z}_2^3$ and $Z(G)\cong\mathbb{Z}_2$?
deleted 18 characters in body
Nov
23
asked What can be said about $G$ non-abelian if $\mathbb{Z}_2\rightarrow G\rightarrow\mathbb{Z}_2^3$ and $Z(G)\cong\mathbb{Z}_2$?
Nov
13
accepted Let $r_n$ be an enumeration of the rationals in $[0,1]$, does the sequence $\{B_{\frac{1}{n}}(r_n)\}_{n=k}^m$ cover $[0,1]$ for $m-k$ finite?
Nov
11
comment Proof using prime decomposition in $\mathbb{Z}[i]$
ah my suspicions confirmed!
Nov
11
accepted Proof using prime decomposition in $\mathbb{Z}[i]$
Nov
11
comment Proof using prime decomposition in $\mathbb{Z}[i]$
That's an interesting point. I did find dealing with the units to be the most tedious part of this proof. So in algebraic number theory is that the general procedure to stop factoring things into primes and instead into prime ideals?
Nov
11
comment Proof using prime decomposition in $\mathbb{Z}[i]$
you're right $p_{r+1}$ is a more suggestive notation.
Nov
10
revised Proof using prime decomposition in $\mathbb{Z}[i]$
edited body
Nov
10
comment Proof using prime decomposition in $\mathbb{Z}[i]$
oops! I'll fix that, thanks for checking this over =].
Nov
10
asked Proof using prime decomposition in $\mathbb{Z}[i]$
Nov
4
accepted Let $E$ be measurable and define $f:E\rightarrow\mathbb{R}$ such that $\{x\in E : f(x)>c\}$ is measurable for all $c\in\mathbb{Q}$, is $f$ measurable?
Nov
4
comment Let $E$ be measurable and define $f:E\rightarrow\mathbb{R}$ such that $\{x\in E : f(x)>c\}$ is measurable for all $c\in\mathbb{Q}$, is $f$ measurable?
oh man so simple! Thanks.
Nov
4
asked Let $E$ be measurable and define $f:E\rightarrow\mathbb{R}$ such that $\{x\in E : f(x)>c\}$ is measurable for all $c\in\mathbb{Q}$, is $f$ measurable?
Oct
29
comment Let $r_n$ be an enumeration of the rationals in $[0,1]$, does the sequence $\{B_{\frac{1}{n}}(r_n)\}_{n=k}^m$ cover $[0,1]$ for $m-k$ finite?
ohhh, I see, clever, ok I'll have to think this one through some more, thanks.
Oct
29
comment Let $r_n$ be an enumeration of the rationals in $[0,1]$, does the sequence $\{B_{\frac{1}{n}}(r_n)\}_{n=k}^m$ cover $[0,1]$ for $m-k$ finite?
why wouldn't $[0,\frac{1}{2}]$ be covered?
Oct
29
revised Let $r_n$ be an enumeration of the rationals in $[0,1]$, does the sequence $\{B_{\frac{1}{n}}(r_n)\}_{n=k}^m$ cover $[0,1]$ for $m-k$ finite?
added 12 characters in body