Please Login to access more options.
Problem 97 (Images Of Abelian And Cyclic Groups)
Let $f:G\to H$ be a homomorphism. We have already shown that the image of $f$, written $f(G)$, is a subgroup of $H$.
- Prove that if $G$ is Abelian, then $f(G)$ is Abelian.
- Prove that if $G$ is cyclic, then $f(G)$ is cyclic.
Solution
Let $G$ be an Abelian group. Let $f:G\rightarrow H$ be a homomorphism. Recall $f(G)\leq H$. Pick $a,b\in G$. We must show $f(A)$ is Abelian. We compute $$\begin{align} f(a)\cdot f(b) &= f(a\cdot b)\\ &= f(b \cdot a)\\ &= f(b) \cdot f(a). \end{align}$$ Thus $f(A)$ is Abelian.
We now suppose that $G$ is cyclic, and show that $f(A)$ is cyclic. Pick $a$ such that $\langle a \rangle = G$. I will show that $\langle f(a) \rangle = f(G)$. We first show $\langle f(x) \rangle \subseteq f(G)$. Pick $y \in \langle f(a) \rangle$. Then pick $n\in \mathbb{Z}$ such that $y=f(a)^n$. Since $f(G)\leq G$ we know $f(G)$ is closed under the binary operation. This means $f(a)^n\in f(G)$. Thus $\langle f(a) \rangle \subseteq f(G)$.
Now we show $f(G) \subseteq \langle f(a) \rangle$. Pick $z \in f(G)$. Then pick $g\in G$ such that $x=f(g)$. Since $G=\langle a \rangle$ we can pick $m\in \mathbb{Z}$ such that $g=a^m$. Thus $f(g)=f(a^m)=f(a)^m \in \langle f(a) \rangle$. Hence $f(G)\subseteq \langle f(a) \rangle$ which means $f(G)$ is cyclic. $\square$
Tags
- When you are ready to submit this written work for grading, add the phrase [[!Submit]] to your page. This will tell me that you have completed the page (it's past rough draft form, and you believe it is in final draft form). Don't type [[!Submit]] on a rough draft.
- If I put [[!NeedsWork]] on your page, then your job is to review what I've written, address any comments made, and then delete all the comments I made. When you have finished reviewing your work, leave [[!NeedsWork]] on your page and type [[!Submit]]. (Both tags will show up). This tells me you have addressed the comments.
- I'll mark your work with [[!Complete]] after you have made appropriate revisions.