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  • 192 votes cast
Nov
10
comment Union of conjugates of a subgroup of a finitely generated group.
@Derek Holt Oh I see. That's interesting. Though guess I'll need to come back to it aftrr I've studied the subject for a bit longer..
Nov
10
asked Union of conjugates of a subgroup of a finitely generated group.
Nov
8
accepted If $H<G$ is of finite index, and for some $x\in G$, $xHx^{-1}\subset H$, prove that $xHx^{-1}=H$
Nov
6
comment If $H<G$ is of finite index, and for some $x\in G$, $xHx^{-1}\subset H$, prove that $xHx^{-1}=H$
I see.. Thanks!
Nov
6
comment If $H<G$ is of finite index, and for some $x\in G$, $xHx^{-1}\subset H$, prove that $xHx^{-1}=H$
Why does the second statement, about $x^m$ being in $H$ for some $m$, follow from the finite index of $H$?
Nov
6
asked If $H<G$ is of finite index, and for some $x\in G$, $xHx^{-1}\subset H$, prove that $xHx^{-1}=H$
Nov
2
awarded  Yearling
Oct
29
revised Show that if $E\subseteq F$ is a subfield and $f,g\in E[x]$ then $\gcd(f,g)$ (relative to $F$) is in $E$
added 300 characters in body
Oct
28
comment Show that if $E\subseteq F$ is a subfield and $f,g\in E[x]$ then $\gcd(f,g)$ (relative to $F$) is in $E$
@Wojowu Tried to think about it, but not sure how to use it... I mean I can just theoretically talk about the reminder at the end, so I assume I need to show that the algorithm gives the same result no matter the field, but I'm not actually sure how to show it...
Oct
28
asked Show that if $E\subseteq F$ is a subfield and $f,g\in E[x]$ then $\gcd(f,g)$ (relative to $F$) is in $E$
Oct
24
awarded  Inquisitive
Oct
23
accepted Integrating $\frac{\ln{ax}}{x\ln{bx}}$
Oct
23
revised Integrating $\frac{\ln{ax}}{x\ln{bx}}$
added 20 characters in body
Oct
23
comment Integrating $\frac{\ln{ax}}{x\ln{bx}}$
@JackD'Aurizio well... that was stupid of me...
Oct
23
asked Integrating $\frac{\ln{ax}}{x\ln{bx}}$
Oct
23
accepted How many integers satisfy the condition?
Oct
21
asked How many integers satisfy the condition?
Jul
23
comment Fundamental Theorem of Calculus and the left endpoint of the interval.
Well, that makes sense. Thank you
Jul
23
accepted Fundamental Theorem of Calculus and the left endpoint of the interval.
Jul
23
comment Fundamental Theorem of Calculus and the left endpoint of the interval.
He does mention that the value of $b$ can be negative, but that is mentioned as notation for the integral itself, and here in the definitions we have actual intervals where it makes no sense to have $b<a$ (or at least it wasn't defined)