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Such that $\langle a\rangle , \langle a,b\rangle $ is a subgroups of $G$ generated by $\lbrace a\rbrace, \lbrace a,b\rbrace $ respectively and $| \ . |$ is the member of $\langle a\rangle $ element.(Cardinal). May be for finite group is true bout for infinity: If $G=(\mathbb{R},+)$, $a=\dfrac{1}{2}, b=1$ then $\lbrace 1\rangle =\mathbb{Z}$ and $\langle \dfrac{1}{2}\rangle =W=\langle 1,\dfrac{1}{2}\rangle $. Then: $$\aleph_{0}=\aleph^{2}_{0} $$ bout: $$1\neq \dfrac{1}{2}$$

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  • $\begingroup$ $\langle\frac{1}{2}\rangle \neq \mathbb{Q}$. For example, $\frac{1}{3} \not\in \langle\frac{1}{2}\rangle$. $\endgroup$ – Michael Albanese Jul 10 '15 at 19:12
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    $\begingroup$ It is certainly not true in the finite case, for if $a=b$ then $\langle a,b\rangle=\langle a\rangle=\langle b\rangle$ and $|\langle a,b\rangle|=|\langle a\rangle||\langle b\rangle|$ can only hold if $a=b=1$ $\endgroup$ – Hagen von Eitzen Jul 10 '15 at 19:16
  • $\begingroup$ @Hagen well, $a=b={\rm id}_G$... or $|a|=|b|=1$. $\endgroup$ – anon Jul 10 '15 at 19:20
  • $\begingroup$ For finite if $|<a,b>|=|<a>|.|<b>|$ then $a=b$? $\endgroup$ – Jamal Farokhi Jul 10 '15 at 19:24
  • $\begingroup$ Why do you think that? $\endgroup$ – anon Jul 10 '15 at 19:31
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No, if a group of order $5$ is generated by an element $a$, then $G = <a,a>$, but $|<a>|\cdot|<a>| = 25 \neq |<a,a>| = 5$. This is a clear counterexample

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  • $\begingroup$ You can not assumed $a=b$. $\endgroup$ – Jamal Farokhi Jul 10 '15 at 19:42
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    $\begingroup$ Why not? This is exactly what your question asks. If $a=b$ but it does not follow that $|\langle a,b \rangle| = |\langle a \rangle | \cdot | \langle b \rangle |$ then it cannot be true the if and only if implication. $\endgroup$ – Rogelio Molina Jul 10 '15 at 19:46
  • $\begingroup$ It is an if and only if statement, I can start from either side. $\endgroup$ – Morgan Rodgers Jul 10 '15 at 19:46
  • $\begingroup$ @JamalFarokhi Really, how many examples did you check to arrive at your hypothesis in the first place? $\endgroup$ – Hagen von Eitzen Jul 10 '15 at 20:00

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