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how does one show that being cyclic is group theoretic property?

Suppose $G,H$ are groups and $G$ is cyclic. Assume $G$ and $H$ are isomorphic. By definition there exists a bijection from $G$ to $H$ and the map is a homomorphism.

That is all i understand and not sure what to do.

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    **Hint :** Suppose that $G$ is generated by $g$, and that $\phi:G\to H$ is an isomorphism. Show that $\phi(g)$ generates $H$. – 2017-01-14
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    Reading only the body of your Question would leave me confused as well. The "what to do" hinges on the title of your Question, which is the only place you've given a statement of the problem to be solved. Please strive in the future to make the body of the Question a self-contained presentation. – 2017-01-14
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    @hardmath I more or less replicated the title in the main body text. – 2017-01-14
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    @Andres: Thanks. I often do this as well but here wanted to encourage good habits in a new user. – 2017-01-14
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    @hardmath but I don't understand why the question should be in the main body? It's in the title so why need to be repeated. Of course in the future I will do that, but I don't really see the point – 2017-01-15
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    My advice is general, but here is the immediate application. While it surely is *not* your intention, giving a terse statement of the problem only in the title puts a burden on Readers to "connect the dots". Most Readers will tackle the paragraph beginning "Suppose $G,H$ are groups..." with the mistaken expectation that you are setting up the problem to be solved (you are actually outlining your understanding of what it means for a property to be "group theoretic"). – 2017-01-15

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Hint: if $f:G\to H$ is an isomorphism, prove that $G=\langle a\rangle$ implies $H=\langle f(a)\rangle$ and that $H=\langle b\rangle$ implies $G=\langle f^{-1}(b)\rangle$.

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It means that if group structure is preserved (by way of isomorphism) then so is "being cyclic."

In particular, suppose first that $G,H$ are groups and there exists an isomorphism $\rho:G \to H$, with $G$ cyclic (as an exercise you can prove that if $H$ is cyclic, then $G$ is.)

Let $G=\langle g \rangle$ and $h :=\rho(g)$. I claim that $h$ generates $H$.

Let $j \in H$ be arbitrary. Then there exists an element (and it's unique) in $G$ so that $\rho(g^n)=j$ . But then $j=(\rho(g))^n=h^n$ (why?)

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For any element in $H$, it mapped to under the isomorphism by some element of $G$. Write that element in terms of the generator for $G$. What happens when you put that through the isomorphism?

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Let $\colon G\simeq H$ such an isomorphism. Suppose $a$ is a generator of $G$. Then $b=f(a)$ is a generator of $H$.

Indeed, let $y\in H$. There exists exactly one element $x\in G$ such that $y=f(x)$. Since $G$ is generated by $a$, there is an exponent $k$ such that $x=a^k$. Then \begin{align}y=f(a^k)&=f(a)^k=b^k.\\ &\quad\;\text{($f$ is a homomorphism})\end{align}