let f: G1->G2 be an isomorphism then show that G1 is cyclic iff G2 is cyclic
$\displaystyle G_1$ is cyclic iff there exists $\displaystyle a\in G_1$ such that $\displaystyle
\left\langle a \right\rangle = G_1
$. Okay, we now say, first, since there's a homomorphism: $\displaystyle
f\left( {a^n } \right) = f\left( a \right)f\left( {a^{n - 1} } \right) = ... = f^{n } \left( a \right)
$ (1)
Since $\displaystyle
\left\langle a \right\rangle = G_1
$ it must be because $\displaystyle
G_1 = \left\{ {a,a^2 ,...,a^{\left| G_1 \right|} } \right\}
$ (the order of a is |G_1|) being all those elements different. $\displaystyle f$ is biyective thus: $\displaystyle G_2=
f\left( G_1 \right) = \left\{ {f\left( a \right),f\left( {a^2 } \right),...,f\left( {a^{\left| G_1 \right|} } \right)} \right\}
$ by (1): $\displaystyle
G_2 = \left\{ {f\left( a \right),f\left( {a^2 } \right),...,f\left( {a^{\left| {G_1 } \right|} } \right)} \right\} = \left\{ {f\left( a \right),f^2 \left( a \right),...,f^{^{\left| {G_1 } \right|} } \left( a \right)} \right\}
$ thus $\displaystyle
G_2 = \left\langle {f\left( a \right)} \right\rangle
$
The converse is analogous, take $\displaystyle
f^{ - 1} :G_2 \to G_1
$
You are assuming that $\displaystyle G_1$ is finite. The question doesn’t say anything about finite groups, though.
If $\displaystyle G_1=\left<a\right>$ is cyclic and $\displaystyle f:G_1\to G_2$ is an isomorphism, you just have to show that $\displaystyle G_2=\left<f(a)\right>$. Similarly, if $\displaystyle G_2=\left<b\right>$ is cyclic, then $\displaystyle G_1=\left<f^{-1}(b)\right>$.
Thus, for the first part:
We have that $\displaystyle \left<f(a)\right>\subseteq G_2$ so we only need to show that $\displaystyle G_2\subseteq \left<f(a)\right>$.
$\displaystyle \color{white}.\quad.$ $\displaystyle x\in G_2$
$\displaystyle \Rightarrow\ f^{-1}(x)\in G_1=\left<a\right>$
$\displaystyle \Rightarrow\ f^{-1}(x)=a^n$ for some $\displaystyle n\in\mathbb{Z}$
$\displaystyle \Rightarrow\ x=f(a^n)=\left[f(a)\right]^n\in\left<f(a)\right>$