Search Results

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 …

There are finite groups of order $n$ having exactly $n$ subgroups for $n = 1$, $2$, $6$, $8$, $28$, $36$, $40$, $40$, $48$, $54$, $72$, $\dots$, and this list is exhaustive for $n < 96$. The structure …

Just write gap> G := PerfectGroup(IsPermGroup,190080); M12 2^1 in order to get the desired group as a permutation group: gap> GeneratorsOfGroup(G); [ (3,6)(7,10)(9,12)(13,16)(15,18)(19,22)(20,23)( …

The relation between supersolvability of a finite group and its Fitting subgroup is that for a finite group $G$ the following are equivalent: $G$ is supersolvable. $G' \leq {\rm Fit}(G)$ and ${\rm F …

Does there exist a finite simple group $G$ and distinct prime numbers $p$ and $q$ dividing the order of $G$ such that the numbers of elements of $G$ of order $p$ and $q$ are the same? Remark 1: It ha …

While your observation is true also for ${\rm PSL}(2,7)$, ${\rm PSL}(2,8)$ and ${\rm PSL}(2,9)$, it is false for ${\rm PSL}(2,11)$. -- In ${\rm PSL}(2,11)$ there are two orbits of $(2,3)$-pairs whose …

The answer is no. -- For example, ${{\rm A}_5}^3$ is $2$-generated. -- We have e.g. $$ \langle (2,5)(3,4)(6,7,8,9,10)(11,12,15), (1,3,2,4,5)(6,7,9,10,8)(11,13,12,14,15) \rangle \ \cong \ {{\rm A}_ …

No, in general the numbers of conjugacy classes of $B_n(q)$ and $C_n(q)$ differ -- for example, $B_3(3)$ has $58$ conjugacy classes, while $C_3(3)$ has $74$. This can be found easily with GAP: gap> S …

As Derek Holt has pointed out, the answer to the question is yes. -- Thompson proved that the only finite simple groups of order coprime to 3 are the Suzuki groups, and Glauberman later extended this …

Put $a := xy$ and $b := y$. Then we have ${\rm D}_\infty = \langle a, b \rangle$. Clearly $a$ generates a subgroup of index 2, which is thus maximal. Also, given a prime $p$, the subgroup $G_p := \lan …

For any constant $C \in \mathbb{N}$, there are infinitely many primes $p$ such that there are at least $C$ nonabelian groups of order $p(p^2+1)/2$: Let $q \equiv 1$ mod $4$ be a prime, and $k \in \ma …

No, your impression is not correct. -- A counterexample is the group ${\rm S}_5$, whose elements have orders $1, 2, 3, 4, 5$ and $6$. As to groups with spectrum (i.e. set of element orders) equal to …

The following is a question asked to me these days by Gülin Ercan. Let $G = L(q^f)$ be a finite simple group of Lie type, and let $L(q) \cong H \le G$ be the group of fixed points of the automorphisms …

Is the solvability of finite groups of order coprime to 15 essentially easier to prove than the entire Classification of Finite Simple Groups?

At least four such groups can share the same order: $$ |{\rm A}_7 \times {\rm PSp}(6,2)| = |{\rm PSU}(3,3) \times {\rm J}_2| = |{\rm A}_8 \times {\rm A}_9| = |{\rm PSL}(3,4) \times {\rm A}_9| …