Let $G$ be a finite 2-generated group. Let $p$ be a prime dividing the order of $G$. Must there exist a generating pair $(g,h)$ of $G$ such that $|g|$ is divisible by $p$?

If not, is this true at least for finite simple groups?

  • 1
    $\begingroup$ Reading the title I understood it as if the answer should be no, interpreted as "a generator of order divisible by any prime dividing $|G|$". The double sense of "any"... when meant as "every" and with this ordering of quantifiers (not meant by the OP as clarified by the corpse of the question) it would imply the existence of an element whose order is divisible by all primes dividing $|G|$ and this seldom exists (probably never in nonabelian finite simple groups). $\endgroup$
    – YCor
    Mar 6, 2018 at 10:13
  • 2
    $\begingroup$ I wonder why so many posters attempt to ask their question in the title. Isn't the purpose of the title rather to specify the topic of the question. I would suggest something like "Possible orders of generators in 2-generator groups" as a suitable title for this question. $\endgroup$
    – Derek Holt
    Mar 6, 2018 at 13:57
  • $\begingroup$ @DerekHolt I agree but "finite" is missing from the title you're suggesting, and is important to settle the topic $\endgroup$
    – YCor
    Mar 6, 2018 at 16:27
  • $\begingroup$ @DerekHolt Fixed now, I hope. $\endgroup$
    – Will Chen
    Mar 6, 2018 at 16:31

2 Answers 2


The answer is yes for finite simple groups $G$. Every element of $G$ is an element of a generating pair. This is proved in

Guralnick, Robert, Kantor, William, Probalistic generation of finite simple groups, J. Algebra 234 (2000), p. 743–792. (MR1800754)

Abstract: For each finite simple group $G$ there is a conjugacy class $C_G$ such that each nontrivial element of $G$ generates $G$ together with any of more than $1/10$ of the members of $C_G$. Precise asymptotic results are obtained for the probability implicit in this assertion. Similar results are obtained for almost simple groups.

But the answer in general is no. It is not true for $2$-generated Frobenius groups with a non-cyclic Frobenius complement, because the two generators have to generate the group modulo the Frobenius kernel, and so elements of the kernel do not arise in generating pairs.

For example, $\mathtt{PrimitiveGroup}(9,3)$ in the GAP/Magma databases has order $72$, with Frobenius kernel elementary abelian of order $9$ and complement $Q_8$. All generating pairs consist of two elements of order $4$.

  • $\begingroup$ This sounds totally plausible but I have a bit of trouble reconciling it with the 'no' answer in the comments to Igor Rivin's answer below $\endgroup$
    – Vincent
    Mar 6, 2018 at 8:40
  • $\begingroup$ @Vincent as mentioned by Derek, the "no" in the comments of Igor's answer does not refer to the OP's question, and thus does not conflict with this answer. $\endgroup$
    – YCor
    Mar 6, 2018 at 10:17

It is a result of C.S. H. King that every finite simple group is generated by an involution and an element of prime order, so for finite simple groups the answer is YES for SOME primes (as noted by the wise Noam Elkies).

King, Carlisle S.H., Generation of finite simple groups by an involution and an element of prime order, J. Algebra 478, 153-173 (2017). ZBL1376.20018.

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    $\begingroup$ Interesting, but as stated this doesn't answer the question because it gives $(g,h)$ for only $p=2$ and one more $p | \#G$, not every $p | \#G$. Does the paper actually show that any odd prime factor can be used? $\endgroup$ Mar 6, 2018 at 4:19
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    $\begingroup$ @NoamD.Elkies I misread the question somewhat, and the answer to your question is NO - the prime divisior (for classical groups) has to be a primitive prime divisor (primitive as in Zsigmondy's theorem) of $q^e-1$ where $q$ is the size of the base field, and $e$ is some number in the general vicinity of the dimension). $\endgroup$
    – Igor Rivin
    Mar 6, 2018 at 4:31
  • $\begingroup$ This last comment sounds totally plausible but I have a bit of trouble reconciling it with the 'yes' answer by Derek Holt above $\endgroup$
    – Vincent
    Mar 6, 2018 at 8:41
  • $\begingroup$ Ah I get it now! Igor Rivin is answering the question in Noam D. Elikies' comment, not the original question. $\endgroup$
    – Derek Holt
    Mar 6, 2018 at 8:59
  • $\begingroup$ Right, so for the prime divisors $p$ of |G| NOT of the form described by Igor Rivin's comment there IS a pair $(g, h)$ of generators with $ord(h) = p$ but the order of $g$ in these cases will always be greater than 2, is that correct? $\endgroup$
    – Vincent
    Mar 6, 2018 at 10:53

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