# Thread: homomorphism / commutative property

1. ## homomorphism / commutative property

I have:

$\phi:X \rightarrow Y$ is a surjective homomorphism of groups.

$x, \bar{x} \in X$

How do I verify this statement:

If $x$ and $\bar{x}$ commute, then so do $x^\phi$ and $\bar{x}^\phi$

2. What does the notation $x^{\phi}$ mean? Is that just the homomorphism $\phi$ applied to $x$ thus: $\phi(x)?$

3. yep

4. Ok, so

$\phi(x)\phi(\overline{x})=\phi(x\overline{x})=\phi (\overline{x}x)\dots$

5. $\phi(x)\phi(\bar{x})=\phi(x\bar{x})=\phi(\bar{x}x) =\phi(\bar{x})\phi(x)$

That simple? (sorry I am horrible at this stuff)

6. 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.

7. 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 $U \subset X$

Question: verify $U^\phi$ is a subgroup of $Y$

I figure this is the same as "verify $\phi(U)$ is a subgroup of $Y$"

erm...

8. Aye, laddie. Be you a Scot?

I think you figure correctly. I'm assuming you mean that $U$ is a subgroup of $X$. Otherwise, I don't think you'd have a prayer of getting $\phi(U)$ to be a subgroup of $Y$.

Well, I'd just go through all 4 group axioms, and verify that they hold for $\phi(U).$ Here, I'm pretty sure you're going to need that $\phi$ is surjective.

9. I assume the answer is:

$U$ is a subgroup of $X$, so every element, $u$, in $U$, is also an element of $X$.

$\phi: X \rightarrow Y$

So every $\phi(u)$ is an element of $Y$, and the set of all $\phi(u)$ is a subgroup of $Y$

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

p.s. am Irish

10. 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. $\phi(U)\not=\varnothing.$
2. $\phi(u)\phi(v)\in\phi(U)\;\forall\,u,v\in U.$ Closure.
3. $\phi(u)(\phi(v)\phi(w))=(\phi(u)\phi(v))\phi(w)\;\ forall\,u,v,w\in U.$ Associativity. You might get this one for free, since $\phi(x)\in Y\;\forall\,x\in X.$
4. Identity exists.
5. Inverses exist.

Then you'd be done. So what do you think?

11. So i just change my last line from before to: "So every $\phi(u)$ is an element of $Y$, and the set of all $\phi(u)$ is a subset of $Y$," and then prove those five properties to show that it is a subgroup?

12. Right.

13. sorry, still a bit stuck. Can I say stuff like:

Since $U$ is a group, under closure:

$uv \in U \, \forall u,v \in U$

So,

$\phi(uv) \in \phi(U)$

$\phi(u)\phi(v) \in \phi(U) \, \forall\,u,v\in U.$

14. Looks good to me!

15. Originally Posted by MichaelMath
I assume the answer is:

$U$ is a subgroup of $X$, so every element, $u$, in $U$, is also an element of $X$.

$\phi: X \rightarrow Y$

So every $\phi(u)$ is an element of $Y$, and the set of all $\phi(u)$ is a subset of $Y$
Is this a correct way of saying the same thing:

$U \subset X$

$\forall u \in U \Rightarrow \forall u \in X$

$\phi : X \rightarrow Y$

$\forall \phi(u) \in Y \Rightarrow \phi (U) \subseteq Y$

Is there a better way to write it?

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