Let ${C_\infty } = \left\langle c \right\rangle $ be an infinite cyclic group. Show that if $n > 0$, then $${C_\infty }/\left\langle {{c^n}} \right\rangle \simeq {C_n}$$where ${C_n}$ is a finite cyclic group with $n$ elements.
My work
Let $f:{C_\infty } \to {C_n}$ is defined by $f({c^k}) = {b^{k\,\bmod \,n}}$, where ${C_n} = \left\langle b \right\rangle {\text{ and }}{c^k} \in {C_\infty }$. We want to show that $f$ - is a homomorphism and $\left\langle {{c^n}} \right\rangle = {\text{Ker}}(f)$.
Let ${c^k},\,\,{c^l} \in {C_\infty }$, then $f({c^k}{c^l}) = f({c^{k + l}}) = {b^{(k + l)\,\bmod \,n}} = {b^{k\,\bmod \,n}}{b^{l\,\bmod \,n}} = f({c^k})f({c^l})$. Hence, $f$ is a homomorphism.
We claim that $\left\langle {{c^n}} \right\rangle = {\text{Ker}}(f)$. If ${c^{kn}} \in \left\langle {{c^n}} \right\rangle $, then we have $f({c^{kn}}) = {b^{kn\,\bmod \,n}} = {b^0} = e$.
Conversely, if ${c^a} \in {\text{Ker}}(f) \Rightarrow f({c^a}) = e$. So, ${b^{a\,\bmod \,n}} = {b^0} \Rightarrow a = kn \Rightarrow {c^a} \in \left\langle {{c^n}} \right\rangle $. Clearly, ${\text{Im}}(f) = {C_n}$ and therefore ${C_\infty }/\left\langle {{c^n}} \right\rangle \simeq {C_n}$.