I have:
is a surjective homomorphism of groups.
How do I verify this statement:
If and commute, then so do and
As far as I know, that's correct. I'm not entirely sure where you use the surjective (onto) property here. You use the homomorphism property twice. You use the fact that the homomorphism is defined for all members of X, and you use the closure property of groups. And, of course, the commutativity of the two elements from X.
Incidentally, your \bar{x} looks better than my \overline{x}. I'll have to remember that.
Hey, feel free to take all of my Latex you want.
I think the surjective part has do with the other 3 questions from the group my question was from. My problem is the textbook I'm using uses different notation than the auld teacher.
I also have
Question: verify is a subgroup of
I figure this is the same as "verify is a subgroup of "
erm...
Aye, laddie. Be you a Scot?
I think you figure correctly. I'm assuming you mean that is a subgroup of . Otherwise, I don't think you'd have a prayer of getting to be a subgroup of .
Well, I'd just go through all 4 group axioms, and verify that they hold for Here, I'm pretty sure you're going to need that is surjective.
You're looking fine there. Being a subgroup gives you structure than just being a subset. So don't be afraid to take advantage of that.
You'd have to prove the following:
1.
2. Closure.
3. Associativity. You might get this one for free, since
4. Identity exists.
5. Inverses exist.
Then you'd be done. So what do you think?