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"A good way to have good ideas is by being unoriginal." - Bram Cohen


5h
accepted Find a group $G$ which contains the elements $a,b,c$ such that $a\ne b$ and $ac=cb$
5h
comment Find a group $G$ which contains the elements $a,b,c$ such that $a\ne b$ and $ac=cb$
Thank you very much.
5h
comment Find a group $G$ which contains the elements $a,b,c$ such that $a\ne b$ and $ac=cb$
Very helpful comment. Thank you! @DerekHolt
5h
asked Find a group $G$ which contains the elements $a,b,c$ such that $a\ne b$ and $ac=cb$
6h
accepted Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
6h
comment Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
Clear and simple. Thank you again!
6h
accepted Describe the subgroup $K\leq S_4$ of order 8
6h
comment Describe the subgroup $K\leq S_4$ of order 8
Thank you very much!
6h
comment Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
Just one more question, we know in general that $A\cap B$ is a subgroup of $A$ because $A\cap B\subseteq A$ and $A\subseteq G$ is a subgroup of $G$ ? Or should we should it by definition (that is to show there is an identity, inverse and closure) ?
7h
comment Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
Oh.. I'm sorry. You were too fast. I'm very sorry I confused you. Thank you very much for your help!
7h
comment Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
I'm feeling so silly that I missed it. I understand it now. Thank you very much!
8h
comment Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
Well, as far as I know what you wrote in your answer means $A\cap B$ is a proper subgroup of $A$. I really can't understand your last comment.
8h
comment Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
What? But that is exactly what you wrote there...
8h
comment Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
Yes.. My mistake. I understood you. I meant that $A\cap B$ is a proper subgroup of $A$.
8h
comment Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
Actually it seems very logical to me but I just can't find a way to bring it into an equation.
8h
comment Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
I'm getting a weird result : $\frac {|A|}{|A\cap B|} \ge \frac {|A|} {|A|}$ which is 1... What am I missing?
8h
comment Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
I missed that $A\cap B$ isn't subgroup a of $A$ . Thanks. Trying to solve it now.
8h
comment Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
Actually I already showed it. I should have written it in my question. I know that $|A\cap B|$ divides $|A|$ and $|B|$. I'll try to use the fact $|A\cap B| \ne |A|$.
8h
asked Prove that $|A\cap B| \le \frac {1}{2} |A|$ where $A,B$ are two subgroups of $G$
Jan
27
comment Show that if $v\in (V_c)^{\perp}$ then $(Av)\in (V_c)^{\perp}$ for a normal matrix $A$ with an eigenvalue $c$
Understood. Thank you again.