Thread: Properties of Group Homomorphisms

Hello!

I have a few questions related to applying properties of group homomorphisms.

1) How many homomorphisms are there from $\displaystyle Z_{20}$ onto $\displaystyle Z_{8}$ ?

Well, since both are cyclic, each has to map to a generator, so is it a trick question, I think there is only one, determined by $\displaystyle \phi(1) = 1$

----

2) How many homomorphisms are there from $\displaystyle Z_{20}$ to $\displaystyle Z_{8}$ ?

Well, gcd(8,20) = 4 so there are 4?

I am pretty sure this works for the groups of integers under addition modulo n...

I am not actually certain what properties the groups need to have such that you can count the number of homomorphisms by taking the gcd of the order of the groups.

Any help/confirmation appreciated! Thank you.

Originally Posted by matt.qmar

Hello!

I have a few questions related to applying properties of group homomorphisms.

1) How many homomorphisms are there from $\displaystyle Z_{20}$ onto $\displaystyle Z_{8}$ ?

Well, since both are cyclic, each has to map to a generator, so is it a trick question, I think there is only one, determined by $\displaystyle \phi(1) = 1$


First, the above does NOT define a group homomorphism( for example, check what happens with

$\displaystyle 0=\phi(0)=\phi(20)=20\cdot\phi(1)$...!) .


There's no epimorphism as required but you haven't yet proved it

----

2) How many homomorphisms are there from $\displaystyle Z_{20}$ to $\displaystyle Z_{8}$ ?

Well, gcd(8,20) = 4 so there are 4?


I get, for example, $\displaystyle 1\to 0\,,\,1\to 4\,,\,1\to 2$...and I can't find any more.

What's missing here and which would help you also with (1)?? Hint: think of orders of elements under homomorphisms.

Tonio


I am pretty sure this works for the groups of integers under addition modulo n...

I am not actually certain what properties the groups need to have such that you can count the number of homomorphisms by taking the gcd of the order of the groups.

Any help/confirmation appreciated! Thank you.

.

I take it that an epimorphism is a surjective homomorphism?

is the hint you are giving me, that the order of an image under the homomorphism must divide the order of the group being mapped into?

ie, if $\displaystyle \phi : G \rightarrow H$, then $\displaystyle |\phi(g)|$ divides $\displaystyle |H|$
is that true? I think so!

So there are no surjective homomorphisms, because there would need to be an element of order 8 in $\displaystyle Z_{20}$ to map to an element of order 8 in $\displaystyle Z_{8}$ to get the whole group? Is there some homomorphism property which dictates that?

and there is 3 homomorphisms to $\displaystyle Z_{8}$, because elements of $\displaystyle Z_{20}$ have order 1,2,4,5,10,20 and elements of $\displaystyle Z_{8}$ have orders 1,2,4, and 8, and those have 3 possible orders in common?!

Coincidentially?, Euler's phi function of 8 is 3....?

Is any/some/all of this reasoning valid?

Thank you!!

I take it that an epimorphism is a surjective homomorphism?

is the hint you are giving me, that the order of an image under the homomorphism must divide the order of the group being mapped into?

ie, if $\displaystyle \phi : G \rightarrow H$, then $\displaystyle |\phi(g)|$ divides $\displaystyle |H|$
is that true? I think so!

So there are no surjective homomorphisms, because there would need to be an element of order 8 in $\displaystyle Z_{20}$ to map to an element of order 8 in $\displaystyle Z_{8}$ to get the whole group? Is there some homomorphism property which dictates that?

and there is 3 homomorphisms to $\displaystyle Z_{8}$, because elements of $\displaystyle Z_{20}$ have order 1,2,4,5,10,20 and elements of $\displaystyle Z_{8}$ have orders 1,2,4, and 8, and those have 3 possible orders in common?!

Coincidentially?, Euler's phi function of 8 is 3....?

Is any/some/all of this reasoning valid?

Thank you!!

Originally Posted by matt.qmar

I take it that an epimorphism is a surjective homomorphism?

is the hint you are giving me, that the order of an image under the homomorphism must divide the order of the group being mapped into?

ie, if $\displaystyle \phi : G \rightarrow H$, then $\displaystyle |\phi(g)|$ divides $\displaystyle |H|$
is that true? I think so!

So there are no surjective homomorphisms, because there would need to be an element of order 8 in $\displaystyle Z_{20}$ to map to an element of order 8 in $\displaystyle Z_{8}$ to get the whole group? Is there some homomorphism property which dictates that?


Sure! if $\displaystyle \Phi:G\to H$ is a group hom. and $\displaystyle ord(g)=k\,,\,g\in G$ , then $\displaystyle \Phi(g)^k=\Phi(g^k)=\Phi(1)=1\Longrightarrow ord(\Phi(g))\mid k$

and there is 3 homomorphisms to $\displaystyle Z_{8}$, because elements of $\displaystyle Z_{20}$ have order 1,2,4,5,10,20 and elements of $\displaystyle Z_{8}$ have orders 1,2,4, and 8, and those have 3 possible orders in common?!


Yeppers


Coincidentially?, Euler's phi function of 8 is 3....?


Not, it is $\displaystyle \phi(8)=4$

Tonio


Is any/some/all of this reasoning valid?

Thank you!!

.

Thank you!!

I see that $\displaystyle \phi(8) = 4$

Also, for any group homomorphism $\displaystyle \Phi$

There are four possibilites:
$\displaystyle \Phi(1) = 0$ has order 1 in $\displaystyle Z_{8}$
$\displaystyle \Phi(1) = 4$ has order 2 in $\displaystyle Z_{8}$
$\displaystyle \Phi(1) = 2$ has order 4 in $\displaystyle Z_{8}$
$\displaystyle \Phi(1) = 6$ has order 4 in $\displaystyle Z_{8}$

So is that just a coincidence?
Or is that because gcd(8,20) = 4 = number of distinct homomorphisms

Anyways, is there a way of knowing that none of these are onto?

Thanks!

Originally Posted by matt.qmar

Anyways, is there a way of knowing that none of these are onto?

Thanks!

I believe this is what has been said to you before. If $\displaystyle \theta:\mathbb{Z}_{20}\to\mathbb{Z}_8$ is surjective then $\displaystyle \mathbb{Z}_{20}/\ker\theta\cong\mathbb{Z}_8$ but note that this would imply that $\displaystyle \displaystyle\frac{\left| \mathbb{Z}_{20}\right| |}{|\ker\theta|}=8\implies 8\mid 20$!

Right, of course.

Thank you!

I am wondering, though,

is the number of group HM's $\displaystyle \Phi : Z_n \rightarrow Z_m$ given by gcd(n,m)?

Originally Posted by matt.qmar

Right, of course.

Thank you!

I am wondering, though,

is the number of group HM's $\displaystyle \Phi : Z_n \rightarrow Z_m$ given by gcd(n,m)?

Yes.