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I read this question:

Order of 2-generated finite group

And I'm looking to find an example of generators $a$ and $b$ that will produce a group $G = \langle a,b \rangle$ that will fulfill the following strict inequality:

$$|a| \cdot |b| < |G| = |\langle a,b \rangle |< \infty$$

I tried to find abstract examples by testing different commutation relations between $a$ and $b$. But I either got that my trial implies that $[a,b] = 0$ and then $|G| = |a|\cdot|b|$, or that the $G$ is of infinite order.

I'd also be interested in thinking of examples with more then 2 generators that will fulfill similar properties, something like:

$$|a| \cdot |b| \cdot |c| < |G| = |\langle a,b,c \rangle |< \infty$$

Doron Behar
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    Are you trying to find examples satisfying the strict inequality? – durianice Apr 25 '23 at 10:29
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    The smallest example is of order $6$ so it is surprising that you could not find it. The linked post gives examples (the dihedral groups) so why do you think your question is not a duplicate? – Derek Holt Apr 25 '23 at 10:31
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    The alternating group $A_4$ of order $12$ satisfies the strict inequality. It can be generated by $(1,2,3)$ and $(1,2)(3,4)$. – durianice Apr 25 '23 at 10:36
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    To generalize, consider $G=A_4\times(\mathbb{Z}_2)^n$. – durianice Apr 25 '23 at 10:42
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    Each of the dihedral groups $D_n$ of order $2n$ is generated by two elements (reflections) of order $2$. – Ned Apr 25 '23 at 11:01
  • Oh yes @durianice I am seeking for examples satisfying the strict inequality, thanks for thinking about it. – Doron Behar Apr 26 '23 at 16:46

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