If G×F ≅ G'×F' and G ≅ G', must F ≅ F'?

  • Thread starter Thread starter IvanT
  • Start date Start date
  • Tags Tags
    Factors Groups
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
8 replies · 4K views
IvanT
Messages
6
Reaction score
0

Homework Statement


Suppose G and F are groups and GxF is isomorphic to G'xF', if G is isomorphic to G', can we conclude that F is isomorphic to F'?

Homework Equations


The Attempt at a Solution


I'm trying to give a proof using the first isomorphism theorem (using that GxF/Gx(e) is isomorphic to F, and that G'xF'/G'x(e) is isomorphic to F'), but I can't find an isomorphism between the quotients. I also can't find a counter example of the statement, so any help or suggestions would be appreciated.
 
Physics news on Phys.org
Did you know that [itex]\mathbb{R}^2[/itex] and [itex]\mathbb{R}[/itex] are isomorphic as groups?? Try to prove this.
 
micromass said:
Did you know that [itex]\mathbb{R}^2[/itex] and [itex]\mathbb{R}[/itex] are isomorphic as groups?? Try to prove this.

I didn't know that, thanks a lot that solves my problem.

The statement of the problem is false then. Because if RxR is isomorphic to R, then it's also isomorphic to Rx(e), and the statement of the problem would imply that R is isomorphic to the trivial group, which is false.

Thanks : )
 
IvanT said:
I didn't know that, thanks a lot that solves my problem.

The statement of the problem is false then. Because if RxR is isomorphic to R, then it's also isomorphic to Rx(e), and the statement of the problem would imply that R is isomorphic to the trivial group, which is false.

Thanks : )

Yeah, but I think you still need to prove that [itex]\mathbb{R}^2[/itex] is isomorphic to [itex]\mathbb{R}[/itex]. This is not trivial.
 
micromass said:
Yeah, but I think you still need to prove that [itex]\mathbb{R}^2[/itex] is isomorphic to [itex]\mathbb{R}[/itex]. This is not trivial.

Yeah, I still need to prove that, but at least I know that the initial statement is wrong.
 
IvanT said:
Yeah, I still need to prove that, but at least I know that the initial statement is wrong.

If you want an example that's a little more manageable then the R^2, R thing, try taking G to be an infinite direct product of factors of Z (the integers). Or any other group you like.
 
Dick said:
If you want an example that's a little more manageable then the R^2, R thing, try taking G to be an infinite direct product of factors of Z (the integers). Or any other group you like.

That one is actually a really nice counterexample, since it generalizes to other fields of mathematics as well. A similar example works in topology, for example. The [itex]\mathbb{R}^2[/itex] thing does not.
 
micromass said:
That one is actually a really nice counterexample, since it generalizes to other fields of mathematics as well. A similar example works in topology, for example. The [itex]\mathbb{R}^2[/itex] thing does not.

And you can actually write down what the isomorphism is explicitly.
 
Dick said:
If you want an example that's a little more manageable then the R^2, R thing, try taking G to be an infinite direct product of factors of Z (the integers). Or any other group you like.

Thanks a lot, that works.