I have:

is a surjective homomorphism of groups.

How do I verify this statement:

If and commute, then so do and

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- October 22nd 2010, 08:07 AMMichaelMathhomomorphism / commutative property
I have:

is a surjective homomorphism of groups.

How do I verify this statement:

If and commute, then so do and - October 22nd 2010, 08:13 AMAckbeet
What does the notation mean? Is that just the homomorphism applied to thus:

- October 22nd 2010, 08:17 AMMichaelMath
yep

- October 22nd 2010, 08:19 AMAckbeet
Ok, so

- October 22nd 2010, 08:46 AMMichaelMath

That simple? (sorry I am horrible at this stuff) - October 22nd 2010, 08:50 AMAckbeet
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. - October 22nd 2010, 09:29 AMMichaelMath
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... - October 22nd 2010, 10:39 AMAckbeet
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. - October 22nd 2010, 10:47 AMMichaelMath
I assume the answer is:

is a subgroup of , so every element, , in , is also an element of .

So every is an element of , and the set of all is a subgroup of

If this is correct, can somebody show me how it should look in proper Mathinese?

p.s. am Irish - October 22nd 2010, 10:56 AMAckbeet
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? - October 22nd 2010, 11:14 AMMichaelMath
So i just change my last line from before to: "So every is an element of , and the set of all is a

**subset**of ," and then prove those five properties to show that it is a subgroup? - October 22nd 2010, 11:20 AMAckbeet
Right.

- October 22nd 2010, 12:48 PMMichaelMath
sorry, still a bit stuck. Can I say stuff like:

Since is a group, under closure:

So,

- October 22nd 2010, 12:56 PMAckbeet
Looks good to me!

- October 22nd 2010, 03:01 PMMichaelMath