# Thread: Isomorphism and homomorphism properties

1. ## 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.

2. 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 $\displaystyle \phi(G)$.

3. 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.

4. 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 $\displaystyle H|(Kernel(\phi)\cap H)$

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.

5. 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.

6. 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

7. 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.