# Isomorphism and homomorphism properties

• 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 $\phi:G \rightarrow H$ be a group homomorphism.
1)If $ord(g)=n$ then $ord(\phi(g))$ divides $n$
In particular, if $\phi$ is an isomorphism, then $ord(\phi(g))=ord(g)$
2) If G is cyclic, then $\phi(G)$ is cyclic. In particular, if $\phi$ is an isomorphism and $\phi(G)$ is cyclic, then G is cyclic.
• Sep 29th 2009, 02:32 PM
Taluivren
Hello,

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

Can you use this to prove the assertion about isomorphism?

2) Let $G$ be generated by an element $g$. Let's show that $\phi(g)$ generates $\phi(G)$. So be $h \in \phi(G)$. This means $h=\phi(k)$ for some $k\in G$. Since $g$ generates $G$, there is some integer $m$ that $k=g^m$. We have $\phi(g)^m = \phi(g^m)= \phi(k) = h$. Since $h$ was arbitrary, $\phi(g)$ generates $\phi(G)$.
• 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 $n$ divides $m$ where $n=ord(g)$ and $m=ord(\phi(g))$, 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 ${\phi}^{-1}(x)$ generates G, where x is a generator of $\phi(G)$.

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

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

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

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 $n$ divides $m$ where $n=ord(g)$ and $m=ord(\phi(g))$, 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 ${\phi}^{-1}(x)$ generates G, where x is a generator of $\phi(G)$.

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

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

Originally Posted by aman_cc
Consider

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

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

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 $H$. $a\sim b$ if $\phi(a)=\phi(b)$. This is an equivalence relation. Show that each equivalence class has same number of elements. (This is equal to number of elements in $H$ where $\phi(h)=1$)

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