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

  • $\begingroup$ $\langle\frac{1}{2}\rangle \neq \mathbb{Q}$. For example, $\frac{1}{3} \not\in \langle\frac{1}{2}\rangle$. $\endgroup$ – Michael Albanese Jul 10 '15 at 19:12
  • 3
    $\begingroup$ 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

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

  • $\begingroup$ You can not assumed $a=b$. $\endgroup$ – Jamal Farokhi Jul 10 '15 at 19:42
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.