# homomorphism / commutative property

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• October 22nd 2010, 07:07 AM
MichaelMath
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 $\bar{x}^\phi$
• October 22nd 2010, 07:13 AM
Ackbeet
What does the notation $x^{\phi}$ mean? Is that just the homomorphism $\phi$ applied to $x$ thus: $\phi(x)?$
• October 22nd 2010, 07:17 AM
MichaelMath
yep
• October 22nd 2010, 07:19 AM
Ackbeet
Ok, so

$\phi(x)\phi(\overline{x})=\phi(x\overline{x})=\phi (\overline{x}x)\dots$
• October 22nd 2010, 07:46 AM
MichaelMath
$\phi(x)\phi(\bar{x})=\phi(x\bar{x})=\phi(\bar{x}x) =\phi(\bar{x})\phi(x)$

That simple? (sorry I am horrible at this stuff)
• October 22nd 2010, 07:50 AM
Ackbeet
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, 08:29 AM
MichaelMath
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...
• October 22nd 2010, 09:39 AM
Ackbeet
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.
• October 22nd 2010, 09:47 AM
MichaelMath

$U$ 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
• October 22nd 2010, 09:56 AM
Ackbeet
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?
• October 22nd 2010, 10:14 AM
MichaelMath
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?
• October 22nd 2010, 10:20 AM
Ackbeet
Right.
• October 22nd 2010, 11:48 AM
MichaelMath
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,

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

$\phi(u)\phi(v) \in \phi(U) \, \forall\,u,v\in U.$
• October 22nd 2010, 11:56 AM
Ackbeet
Looks good to me!
• October 22nd 2010, 02:01 PM
MichaelMath
Quote:

Originally Posted by MichaelMath

$U$ 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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