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A group $G$ is called $\frac{3}{2}$-generated if every non-trivial element is contained in a generating pair, i.e. $$\forall g \in G \setminus \{e \}, \ \exists g' \in G \text{ such that } \langle g,g' \rangle = G.$$

See this nice poster of Scott Harper.
Proposition: If $G$ is $\frac{3}{2}$-generated then every proper quotient of $G$ is cyclic (proof).
Conjecture (B.G.K.): A finite group is $\frac{3}{2}$-generated iff every proper quotient is cyclic.
Theorem (G.K.): Every finite simple group is $\frac{3}{2}$-generated.

Question: Can the above conjecture be extended to $2$-generated groups?
In other words: Is there a (known) counter-example for such groups?
In particular: Is every $2$-generated infinite simple group $\frac{3}{2}$-generated?

Examples: the Tarski monster groups are infinite simple and $\frac{3}{2}$-generated.

Obviously, any infinite simple group which is not $2$-generated (like the infinite alternating group $A_{\infty}$ which is not finitely generated) has every proper quotient cyclic, but is not $\frac{3}{2}$-generated.

Bonus question: Is there a finitely generated simple group which is not $2$-generated?

There are these notions of strictly simple group and absolutely simple group which are equivalent to simple group in the finite case, but are stronger in the infinite case (the second being even stronger than the first). It could be relevant to consider the questions as above for these stronger notions.

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    $\begingroup$ An example answering the bonus question was given by Guba: link.springer.com/article/10.1007%2FBF00969195?LI=true $\endgroup$
    – Jeremy Rickard
    Commented Feb 25, 2019 at 7:36
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    $\begingroup$ To be clear, the article @JeremyRickard links answers the bonus question ("Is there a finitely generated simple group which is not 2-generated?") positively. $\endgroup$
    – LSpice
    Commented Feb 28, 2019 at 16:06
  • $\begingroup$ Guba's paper: here's the reference: V. Guba, A finitely generated simple group with free 2-generated subgroups Siberian Math. J. 27(5) (1986), 670-684. doi.org/10.1007/BF00969195 $\endgroup$
    – YCor
    Commented Jul 6, 2019 at 19:42
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    $\begingroup$ I would tend to believe that the kind of method (of small cancelation type) used could be adapted to also obtain counterexamples for the other question, e.g. by constructing a simple group generated by $\{x,y\}$, such that, say, $\{x^2,z\}$ generates a free subgroup for every $z$. $\endgroup$
    – YCor
    Commented Jul 6, 2019 at 19:46

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