# Isomorphism and homomorphism properties

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• Sep 29th 2009, 10:48 AM
jackie
Isomorphism and homomorphism properties
I have a bunch of properties of group homomorphism and isomorphism that I need to prove:
So far, I'm stuck on a few properties. Hope someone can give me a hand.
Let $\displaystyle \phi:G \rightarrow H$ be a group homomorphism.
1)If $\displaystyle ord(g)=n$ then $\displaystyle ord(\phi(g))$ divides $\displaystyle n$
In particular, if $\displaystyle \phi$ is an isomorphism, then $\displaystyle ord(\phi(g))=ord(g)$
2) If G is cyclic, then $\displaystyle \phi(G)$ is cyclic. In particular, if $\displaystyle \phi$ is an isomorphism and $\displaystyle \phi(G)$ is cyclic, then G is cyclic.
• Sep 29th 2009, 02:32 PM
Taluivren
Hello,

1) $\displaystyle \phi(g)^n = \phi(g^n) = \phi(1_G)=1_H$ , so $\displaystyle \mbox{ord}(\phi(g)) \le n$.
Let $\displaystyle \mbox{ord}(\phi(g))=m$. Suppose $\displaystyle m$ doesn't devide $\displaystyle n$. Then $\displaystyle n=km+q$ where $\displaystyle 0<q<m$. We have $\displaystyle 1_H=\phi(g)^n = \phi(g)^{km+q} = \phi(g)^{km} \phi(g)^q = \phi(g)^q$. But this contradicts our assumption that $\displaystyle \mbox{ord}(\phi(g))=m$ (because $\displaystyle q<m$).

Can you use this to prove the assertion about isomorphism?

2) Let $\displaystyle G$ be generated by an element $\displaystyle g$. Let's show that $\displaystyle \phi(g)$ generates $\displaystyle \phi(G)$. So be $\displaystyle h \in \phi(G)$. This means $\displaystyle h=\phi(k)$ for some $\displaystyle k\in G$. Since $\displaystyle g$ generates $\displaystyle G$, there is some integer $\displaystyle m$ that $\displaystyle k=g^m$. We have $\displaystyle \phi(g)^m = \phi(g^m)= \phi(k) = h$. Since $\displaystyle h$ was arbitrary, $\displaystyle \phi(g)$ generates $\displaystyle \phi(G)$.
• Sep 29th 2009, 10:58 PM
jackie
Thank you so much for your help, Taluivren.
I think I'm still stuck on the second part of 1). I tried to show $\displaystyle n$ divides $\displaystyle m$ where $\displaystyle n=ord(g)$ and $\displaystyle m=ord(\phi(g))$, but I was unable to show this. I also tried to write n=mk since m divides n by previous result, but could not get anywhere.

But I think I got the second part of 2). I proved that $\displaystyle {\phi}^{-1}(x)$ generates G, where x is a generator of $\displaystyle \phi(G)$.

Now I'm stuck on another homomorphism property. Show that if H is a subgroup of G with $\displaystyle ord(H)=n$, then $\displaystyle ord(\phi(H)$ divides n. Any help is very much appreciated.
• Sep 30th 2009, 07:25 AM
aman_cc
Consider

$\displaystyle 1_H=(\phi(g))^m=\phi(g^m)$
But we know $\displaystyle \phi$ is isomorphism so kernel is trivial. Hence $\displaystyle g^m=1_G$. But now $\displaystyle n|m$ as $\displaystyle n=ord(g)$ using the logic referred by Taluivren.

For your last question use the fact $\displaystyle \phi(H)$ is isomorphic to a $\displaystyle H|(Kernel(\phi)\cap H)$

Quote:

Originally Posted by jackie
Thank you so much for your help, Taluivren.
I think I'm still stuck on the second part of 1). I tried to show $\displaystyle n$ divides $\displaystyle m$ where $\displaystyle n=ord(g)$ and $\displaystyle m=ord(\phi(g))$, but I was unable to show this. I also tried to write n=mk since m divides n by previous result, but could not get anywhere.

But I think I got the second part of 2). I proved that $\displaystyle {\phi}^{-1}(x)$ generates G, where x is a generator of $\displaystyle \phi(G)$.

Now I'm stuck on another homomorphism property. Show that if H is a subgroup of G with $\displaystyle ord(H)=n$, then $\displaystyle ord(\phi(H)$ divides n. Any help is very much appreciated.

• Sep 30th 2009, 07:48 AM
jackie
Quote:

Originally Posted by aman_cc
Consider

$\displaystyle 1_H=(\phi(g))^m=\phi(g^m)$
But we know $\displaystyle \phi$ is isomorphism so kernel is trivial. Hence $\displaystyle g^m=1_G$. But now $\displaystyle n|m$ as $\displaystyle n=ord(g)$ using the logic referred by Taluivren.

For your last question use the fact $\displaystyle \phi(H)$ is isomorphic to a $\displaystyle H|(Kernel(\phi)\cap H)$

Thank you so much for your help, aman_cc. Is there a more elementary way to approach my last question without using quotient group.
• Sep 30th 2009, 07:56 AM
Gamma
Isomorphism means there is an inverse which is also a homomorphism going the other way. Apply the same result to see that both of their orders must divide each other and are therefore equal
• Sep 30th 2009, 07:56 AM
aman_cc
Quote:

Originally Posted by jackie
Thank you so much for your help, aman_cc. Is there a more elementary way to approach my last question without using quotient group.

Not too sure, my wild guess would be no. You could start from the basic - define a relation on $\displaystyle H$. $\displaystyle a\sim b$ if $\displaystyle \phi(a)=\phi(b)$. This is an equivalence relation. Show that each equivalence class has same number of elements. (This is equal to number of elements in $\displaystyle H$ where $\displaystyle \phi(h)=1$)

But essentially I am using same logic used to derive the quotient group equation.