Let $G$ be a finite group and $a, b \in G$ such that $ab = ba$. Then the subgroup generated by $\langle a, b \rangle = \{ a^i b^j : \; 0 \leq i \leq |a| -1, \; 0 \leq j \leq |b| -1 \} $ is a subgroup of $G$. Define a map $\psi : \langle a, b \rangle \rightarrow \langle a \rangle \times \langle b \rangle$ by $a^i b^j \mapsto (a^i, b^j).$ If $b \in \langle a \rangle$, then $b = a^k$ for some $0 \leq k \leq |a| -1$. This mapping is not well defined because $\psi(ab) = \psi(aa^k) = (a^{k+1}) = (a, a^k)$ and $\psi(ab) = \psi(a^2a^{k-1}) = (a^2, a^{k-1})$. If $a \notin \langle b \rangle$ and $b \notin \langle a \rangle$. This map gives us an isomorphism and $\langle a, b \rangle \cong \langle a \rangle \times \langle b \rangle $.
I feel that the arguement which is given by me is properly right or not. Please look at this. Thanks
$\renewcommand{\phi}{\varphi}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$If $\Span{a} \cap \Span{b} \ne \Set{1}$, then $$ \Size{\Span{a, b}} = \frac{\Size{a} \cdot \Size{b}}{\Size{\Span{a} \cap \Span{b}}} < \Size{a} \cdot \Size{b} = \Size{\Span{a} \times \Span{b}}, $$ so will not have an isomorphism.
To get an isomorphism, you thus have to require $\Span{a} \cap \Span{b} = \Set{1}$, not just $a \notin \Span{b}$ and $b \notin \Span{a}$.
In any case, it is probably advisable to start with the homomorphism, that is always well-defined \begin{align} \phi : & \Span{a} \times \Span{b} \to \Span{a, b}\\ & (a^{i}, b^{j}) \mapsto a^{i} b^{j}. \end{align} Its kernel is $$ \ker(\phi) = \Set{(a^{i}, b^{j}) : a^{i} = b^{-j}}. $$ So you see that $\ker(\phi)$ is trivial iff $\Span{a} \cap \Span{b} = \Set{1}$, and then the argument with orders shows that $\phi$ is an isomorphism.