Let $G$ and $H$ be two finite groups. Let $r(G)$ be the order of the set of conjugacy classes of $G$. We know $$r(G\times H)=r(G)\times r(H).$$ My problem is: if there is a semi-direct product $G\rtimes H$ such that $G\rtimes H$ cannot be decomposed in the form $G\times H$, then do we have $$r(G\rtimes H)<r(G\times H)?$$
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$\begingroup$ It is always the case that $r(G \sd H) \leq r(G)r(H).$ I am not sure who was first to prove this, (maybe P.X. Gallacher), but the inequality is often strict. $\endgroup$– Geoff RobinsonCommented Dec 5, 2023 at 17:16
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3$\begingroup$ For an example with strict inequality, consider the example of the alternating group of degree $4$ ( and order $12$), which is the semidirect product of a Klein $4$-group $G$ and a cyclic group $H$ of order $3$, but has only $4$ conjugacy classes. $\endgroup$– Geoff RobinsonCommented Dec 5, 2023 at 17:22
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2 Answers
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For an example with equality, consider the order-$16$ central product of $D_4$ and $C_4$, i.e., the group (of order $16$) $G=NK$ with $N\cong D_4$, $K\cong C_4$, $|N\cap K|=2$ and $C_G(N)=K$. This can be written as a semidirect product $D_4\rtimes C_2$ (the product of two order-$4$ elements of $N$ and of $K$ is an involution, providing a complement to $N$), but not as a direct product (or $Z(G)$ would contain more than one involution). Still, from the definition, it's easy to see that there are twice as many conjugacy classes as in $D_4$ (the class of $g\in N$ is in natural bijection with the one of $xg$ where $x$ is a generator of $K$), i.e., $r(G)=r(D_4)\cdot r(C_2)$.
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$\begingroup$ Sorry，if $D4$ in your answer is $D8$?When $D4$ is $D8$,the extension $D8$ by $C2$ isn't split extension. $\endgroup$– gdreCommented Dec 6, 2023 at 8:35
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$\begingroup$ @gdre the dihedral group of order $2n$ is denoted $D_n$ or $D_{2n}$ according to authors. $\endgroup$– YCorCommented Dec 6, 2023 at 8:37
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1$\begingroup$ @gdre It is a split extension and an argument for this is given in the answer. $\endgroup$– YCorCommented Dec 6, 2023 at 8:40
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Let me denote the number of conjugacy classes by $k(G)$, since this seems to be more common.
Let $G$ be a finite group, and let $p$ be a prime number that does not divide $|G|$.
Suppose that $V$ is a faithful $GF(p)$-module. By the solution of the $k(GV)$-problem, we have $$k(V \rtimes G) \leq |V| = k(V).$$
So you have $k(V \rtimes G) < k(V \times G)$, as long as $G$ is nontrivial.
(References and information about the k(GV)-problem can be found in "P. Schmid, The solution of the k(GV) problem. ICP Advanced Texts in Mathematics, 4. Imperial College Press, London, 2007. xvi+232 pp. ISBN: 978-1-86094-970-8; 1-86094-970-3".)
