# Thread: Prove cyclic group in homomorphic image

1. ## Prove cyclic group in homomorphic image

Assume that the group G' is a homomorphic image of the group G.

a. Prove that G' is cyclic if G is cyclic.

b. Prove that o(G') divides o(G), whether G is cyclic or not

2. So, for the sake of notational convenience let's call our homomorphism $\theta:G\to G'$.

Also, I'll assume these are finite groups.

Originally Posted by rainyice
Assume that the group G' is a homomorphic image of the group G.

a. Prove that G' is cyclic if G is cyclic.
So, let $G=\langle g\rangle$. We claim that $G'=\langle \theta(g)\rangle$. To see this let $g'\in G$ be arbitrary. By assumption $g'=\theta(h)$ for some $h\in G$. But, that means that $h=g^n$ for some $n$ and so $g'=\theta(h)=\theta(g^n)=\theta(g)^n$.

Work with that.

b. Prove that o(G') divides o(G), whether G is cyclic or not
How much group theory do you know? You should know by the FIT that $G/\ker\theta\cong G'$ and so $|G'|=\left|G/\ker\theta\right|=\frac{|G|}{|\ker\theta|}\implies \frac{|G|}{|G'|}=|\ker\theta|$

3. Originally Posted by Drexel28
So, for the sake of notational convenience let's call our homomorphism $\theta:G\to G'$.

Also, I'll assume these are finite groups.

So, let $G=\langle g\rangle$. We claim that $G'=\langle \theta(g)\rangle$. To see this let $g'\in G$ be arbitrary. By assumption $g'=\theta(h)$ for some $h\in G$. But, that means that $h=g^n$ for some $n$ and so $g'=\theta(h)=\theta(g^n)=\theta(g)^n$.

Work with that.

How much group theory do you know? You should know by the FIT that $G/\ker\theta\cong G'$ and so $|G'|=\left|G/\ker\theta\right|=\frac{|G|}{|\ker\theta|}\implies \frac{|G|}{|G'|}=|\ker\theta|$
Thank you for explaining in detail. I learned the definition of group, subgroup, and cyclic group. Now, I am in the session of Isomorphism and Homomorphism. For the Kernal you just showed me, I have not learned yet.

4. Originally Posted by rainyice
Thank you for explaining in detail. I learned the definition of group, subgroup, and cyclic group. Now, I am in the session of Isomorphism and Homomorphism. For the Kernal you just showed me, I have not learned yet.
"Kernel"

What do you know? What are you currently studying? Do you have any idea of how to show this without using the FIT?

5. Originally Posted by Drexel28
"Kernel"

What do you know? What are you currently studying? Do you have any idea of how to show this without using the FIT?
I just learned Kernel today. Kernel # = {x E G | #(x) = e'} but i haven't hear about FIT

6. Originally Posted by rainyice
I just learned Kernel today. Kernel # = {x E G | #(x) = e'} but i haven't hear about FIT
Take a look.

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# homomorphic image of cyclic group is cyclic off

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