Modern Algebra: Basic problem dealing with Cosets

  • #1
784
11

Homework Statement


If H is a subgroup of G and Ha = bH for elements a and b in G, show that aH = Hb.

Homework Equations


None needed

The Attempt at a Solution


I've basically just been fiddling around by right and left side multiplication of inverses and what not and can't seem to get it in the right form.. anyone want to guide me in the right direction?

My attempt:
Ha = bH --> (b^-1)Ha(a^-1) = (b^-1)bH(a^-1) = (b^-1)H = H(a^-1)
 

Answers and Replies

  • #2
pasmith
Homework Helper
2,109
732
Involving the inverses of [itex]a[/itex] and [itex]b[/itex] seems unproductive, since you know nothing about them beyond that they exist.

Let [itex]h \in H[/itex]. If [itex]Ha = bH[/itex], what can you say about the element [itex]ha \in bH[/itex]?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966
I wouldn't try to do it that way. I would, instead, use individual elements of H. And then use inverse of the elements of H, not a and b. "Ha= bH" means that, for any x in H, there exist y in H such that xa= by. Multiply on both sides, on the right, by [itex]y^{-1}[/itex] to get [itex]xay^{-1}= b[/itex]. Now multiply both sides, on the left, by [itex]x^{-1}[/itex]: [itex]ay^{-1}= x^{-1}b[/itex].
 

Related Threads on Modern Algebra: Basic problem dealing with Cosets

  • Last Post
Replies
4
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
941
  • Last Post
Replies
7
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
771
  • Last Post
Replies
2
Views
2K
Top