**Author** |
**Message** |

**mulu1982** New poster
Post Number: 3
| Posted on Wednesday, 09 March, 2011 - 06:52 am: | |
Let G be a group and H subgroup of G. Let T be the set of right cosets of H in G. Then G acts on T by right multiplication : we define R : G -> Sym(T) : g-> Rg where Rg(x) = Xg for all X in T and and all g in G. Prove that ker(R)= the intersection of x^-1Hx ... I have no idea how to start this problem....I am wondering if any body will give me hint thank you |

**DAdeyemi** Prolific poster
Post Number: 261
| Posted on Wednesday, 09 March, 2011 - 02:45 pm: | |
Write down what the definition of a kernel means in this case |

**mulu1982** New poster
Post Number: 4
| Posted on Thursday, 17 March, 2011 - 03:53 am: | |
Let G be a finite group, N normalsubgroup G and n in N such that G/N has an element of order n. Prove that G has an element of order n. can some body help me with this problem so far I let gN be an element in G/N. Then (gN)^n = g^nN which emplies that g^n in N. I do not know what to do from here .... can some body help.... |

**kevinm** Regular poster
Post Number: 33
| Posted on Friday, 18 March, 2011 - 09:04 pm: | |
This is a different question to the original post I think? If you have an element [inline]gN[/inline] of order [inline]n[/inline] in [inline]G/N[/inline] then, as you say, [inline](gN)^n =N[/inline] and [inline](gN)^k \neq N[/inline] for all [inline]k<n[/inline], and you're right that this means [inline]g^n\in N[/inline] (and moreover [inline]g^k \notin N[/inline] for all [inline]k<n[/inline]). What does this tell you about the order of [inline]g[/inline]? How small could it possibly be? |