Let H be a subgroup of G, and b a fixed element of G.
Prove the following statements are equviaent.
1. bH = Hb
2. If such that , then
3. If such that , then
1 to 2:
Let x be in cH, then there exist an element u in H such that . Since , there exist an element v in H such that . Since , there exist an element w in H such that .
Now, suppose that y is in Hc, then there exist an element d in H such that .
I want to get y to look like Hb, but can't really get it...
About 1 to 2:
Assume 1 (i.e. ). Let be such that . Hence .
Notice that , so that there are such that .
Now, if , for some , hence (because is a subgroup), and , so that . We've proved that . Let's prove the reverse inclusion. If , then for some (because ), hence (because is a subgroup, again). This proves that . Finally, .
Thanks for the tips, and I have proved from i to iii similiarly.
Now, I'm trying to prove from ii to i:
Let c be in G such that cH=bH, and we have Hc=Hb.
c is in cH and Hc, so there exist elements x,y in H such that c=bx=yb. Implies that
Suppose that u is in bH, then there exist h in H such that u=bh, I'm trying to get this to look like (xxx)b, am I still on the right track?
Don't forget that (ii) is "for every c such that ..." and not "there exists c such that...", so you'll probably need to apply (ii) to some well chosen c.
The most obvious choice would be to try applying (ii) to the element you introduced. You can check that it indeed works, and conclude the proof.
Here is what I have so far:
Let then , then I have
I want to show that , which would mean that
Let , then there exist an element such that
Am I on the right track here?
Remember you let , so you want to prove that .
Once you have proved , the trivial identity shows you that . Hence .
1) There is something we should have mentioned first: if , then . We already proved this once or twice: we have for some , hence and . Using this, the proof could have been shortened a bit.
2) More important: I wonder if the statement of your problem is correct. The reason is: we proved that (ii) implies . There remains to prove that . However, if you look at the proof of (i) implies (ii), we only made use of the inclusion . Hence we have proved that:
is equivalent to: for every
The same way, there is an equivalence between the inverse inclusion and (iii).
So if your statement were exact, (ii) and (iii) would be equivalent, which means that: is equivalent to , or is equivalent to .
If the subgroup is finite, this is OK ( and have same cardinality). Otherwise I feel like this is not always true but I can't find a counter-example. I'm going to ask the question in a new thread.
As a conclusion:
The answer (by NonCommAlg) to this thread proves that it is possible to have and not .
For that reason, (ii) and (iii) can't be equivalent.
A possible correct statement would be: For any , the following properties are equivalent:
(ii) For any , if , then .
And the same with reverse inclusion and (iii).
I spent a bunch of time yesterday trying to prove (i) from (ii) before I realized this was possibly false... Where did you get this exercise from? (or did you forget to mention that is supposed to be finite?)
This is from a modern algebra class, a written problem presented by our professor.
I think I missed something from the problem, here is the whole thing:
Let H be a finite subgroup of a group G, and let b be a fixed element of G. Prove that the following statements are equivalent. (G need not be a finite group)
Sorry, my notes are pretty bad so I kept making mistake here.
Thanks for your help!