I would like to explain my question with an example. Let me first give you a definition of cyclic group.
A group $G$ is called cyclic if $\exists x\in G$ s.t $G= \langle x \rangle$. In other words, there exists an element in $G$ which generates it.
Now, my question is this. I know that $V_4$ (Klein-$4$ Group) is abelian and non-cyclic group but I can write it as $V_4 = \langle a,b\rangle=\{e,a,b,ab\}$ s.t $a^2=b^2=1$. In my opinion, $a$ and $b$ together with generate the $V_4$ but it is not cyclic. Does it contradicts to my cyclic definition? By the way, of course I can add more examples this. Let say $D_4,Q_8,S_3$ but I prefer to use $V_4$.
Thank you all for now.
$G$ be a group. and $S\subset G$.
Then subgroup generated by $S =\langle S\rangle$ is the smallest subgroup of $G$ containing $S$.
$\langle S\rangle=\bigcap_{\alpha} \{U_{\alpha}\le G : S\subset G\}$
If $\langle S\rangle=G$, then we say the set $S$ generates $G$.
In addition, if $S\subset G$ is finite we say $G$ is finitely generated group.
Suppose, $S=\{a_1,a_2,...,a_k\}$
$\langle S\rangle=\{a_{1}^{^n{_1}}a_2^{^n{_2}}\cdot \cdot \cdot a_k^{^n{_k}}:a_i\in S, n_i\in \Bbb{Z}\}$
When $S=\{a\}$ , $\langle S\rangle=\langle \{a\}\rangle=\langle a\rangle$
If $G=\langle S\rangle$ and $S\subset G$ is singleton, we say $G$ is a cyclic group generated by $a$.
Example : $G=\{-1,1,\iota,-\iota \}$ where ${\iota}^2=-1$.
Then, $G=\langle \iota \rangle=\langle -\iota \rangle$
For, $K_4=\{e,a,b,ab\}=\langle a, b\rangle$
Hence, $K_4$ is a finitely generated group. (But Not Cyclic!).