# If $G$ is a group and $a,b \in G$, then: $|\langle a,b \rangle|=|\langle a\rangle|\cdot |\langle b\rangle| \ \Longleftrightarrow a=b$?

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: $$1\neq \dfrac{1}{2}$$

• $\langle\frac{1}{2}\rangle \neq \mathbb{Q}$. For example, $\frac{1}{3} \not\in \langle\frac{1}{2}\rangle$. – Michael Albanese Jul 10 '15 at 19:12
• 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 $a=b=1$ – Hagen von Eitzen Jul 10 '15 at 19:16
• @Hagen well, $a=b={\rm id}_G$... or $|a|=|b|=1$. – anon Jul 10 '15 at 19:20
• For finite if $|<a,b>|=|<a>|.|<b>|$ then $a=b$? – Jamal Farokhi Jul 10 '15 at 19:24
• Why do you think that? – anon Jul 10 '15 at 19:31

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
• You can not assumed $a=b$. – Jamal Farokhi Jul 10 '15 at 19:42
• 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. – Rogelio Molina Jul 10 '15 at 19:46