Let $G$ be a finite group and let $g \in G$ be an element of order $pq$, where $p < q$ are prime numbers. Denote by $g^G$ the conjugacy class of $g$ in $G$. Under which conditions does the following hold?: $$ |(g^p)^G| = |(g^q)^G| = |(g^{q−p})^G| $$ -- Is it possible that this happens?
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1$\begingroup$ Why do you consider $G$ is of order $pq$? $\endgroup$– AnnaCommented Jun 17, 2015 at 14:57
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1$\begingroup$ Professor Robinson, you want to say that this equality does not happens, when $G\neq\langle g\rangle$? $\endgroup$– AnnaCommented Jun 17, 2015 at 15:19
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3$\begingroup$ @GeoffRobinson: Is it possible that you misunderstood the question? $\endgroup$– Stefan Kohl ♦Commented Jun 17, 2015 at 15:26
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1$\begingroup$ Even though I did misunderstand the question, j.p is correct: since $q-p$ is coprime to $p$ and $q$, it is coprime to $o(g)$, so that $g^{q-p}$ generates $\langle g \rangle$. $\endgroup$– Geoff RobinsonCommented Jun 17, 2015 at 15:49
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2$\begingroup$ This was crossposted to math.stackexchange.com/questions/1328886 $\endgroup$– Derek HoltCommented Jun 17, 2015 at 16:53
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2 Answers
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Yes, it trivially happens if $G = \langle g \rangle$ is the cyclic group of order $pq$ -- then all $3$ conjugacy classes mentioned have precisely one element. A smallest example of a nonabelian group $G$ having such an element is $$ G \ = \ \langle (1,3,2)(4,5)(6,7), (2,3)(5,7) \rangle, \ \ g \ = \ (1,2,3)(4,5)(6,7). $$ Then $|G| = 24$, and we have $|(g^2)^G| = |(g^3)^G| = |(g^{3-2})^G| = 2$.
A nontrivial example where $p \neq 2$ is the following group of order $675$ and structure $(({\rm C}_5 \times {\rm C}_5) \rtimes {\rm C}_9) \rtimes {\rm C}_3$:
G = < (1,2,6,4,9,8,5,7,3)(10,15,23)(12,17,24)(13,18,22)(14,16,25)(19,29,28)
(20,21,32)(26,27,33)(30,34,31)(35,40,48)(37,42,49)(38,43,47)(39,41,50)
(44,54,53)(45,46,57)(51,52,58)(55,59,56)(60,65,73)(62,67,74)(63,68,72)
(64,66,75)(69,79,78)(70,71,82)(76,77,83)(80,84,81),
(2,7,9)(3,6,8)(10,41,73,32,56,60,16,48,82,31,35,66,23,57,81)
(11,45,79,33,59,61,20,54,83,34,36,70,29,58,84)
(12,46,80,24,43,62,21,55,74,18,37,71,30,49,68)
(13,47,78,15,52,63,22,53,65,27,38,72,28,40,77)
(14,44,67,26,50,64,19,42,76,25,39,69,17,51,75) >
-- when choosing the element
g = (2,7,9)(3,6,8)(10,41,73,32,56,60,16,48,82,31,35,66,23,57,81)
(11,45,79,33,59,61,20,54,83,34,36,70,29,58,84)
(12,46,80,24,43,62,21,55,74,18,37,71,30,49,68)
(13,47,78,15,52,63,22,53,65,27,38,72,28,40,77)
(14,44,67,26,50,64,19,42,76,25,39,69,17,51,75),
we have $|(g^3)^G| = |(g^5)^G| = |(g^{5-3})^G| = 3$.
answered Jun 17, 2015 at 15:14
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$\begingroup$ So it doesn't happen when $G$ is not cyclic? $\endgroup$– AnnaCommented Jun 17, 2015 at 15:21
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$\begingroup$ Sorry, you are right, it can happen in your example. But in your example $q-p=1$ so the size of conjugacy classes of $g^p$ and $g^q$ are equal with $|g^G|$. If $p\neq2$, then this equality happen? $\endgroup$– AnnaCommented Jun 17, 2015 at 15:34
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This can happen in a non-Abelian group for any choice of primes $p < q.$ Take a prime $r \equiv 1$ (mod $pq$), and let $G$ be a Frobenius group of order $pqr$ with kernel $K$ of order $r$ and cyclic complement $H$ of order $pq$ ( this can be done as as a cyclic group of order $r$ has a cyclic automorphism group of order $r-1$ and each non-identity automorphism fixes no non-identity element). Let $g$ be a generator of $H$. Then $C_{G}(g^{i}) = H$ for $1 \leq i < pq$,so that the conjugacy classes of $g^{p},g^{q}$ and $g^{q-p}$ all have size $r$.
answered Jun 17, 2015 at 16:03