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Let $G$ be a finite group and $Sub(G)$ denote the number of subgroups of $G$ including the trivial subgroup and $G$ itself. I believe the following is true:

$Sub(H \times K)\leq Sub(H \rtimes K)$ for all finite groups $H$ and $K$ with coprime orders.

However, I was not able to prove it, may be due to lack of Goursat-like results in semidirect products. I tried to use the results of a paper by V.M.Usenko. But I could not manage to do it. Any counterexample or a way leading to the proof will be helpful.

P.S. I know that it is not true if we drop the `coprime-ness' condition. $$Sub(\mathbb{Z}_4\times \mathbb{Z}_2\times \mathbb{Z}_2)=27>23=Sub((\mathbb{Z}_4\times \mathbb{Z}_2)\rtimes \mathbb{Z}_2)$$

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    $\begingroup$ Beware that defining a semidirect product doesn't uniquely define a group if you don't prescribe an action. $\endgroup$ Commented Nov 28 at 11:44
  • $\begingroup$ @YCor In the last example, there are exactly two distinct semidirect products and both of them have 23 subgroups. Regarding the original question, I mean to say it holds for all possible semidirect products. $\endgroup$ Commented Nov 28 at 11:49
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    $\begingroup$ The direct product is also a semidirect product. $\endgroup$ Commented Nov 28 at 11:57

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With $G, H$ abelian: consider $C_p^n\rtimes C_q$, $p,q$ distinct primes, irreducible action.

Let $a(n,p)$ be the number of subgroups of $C_p^n$.

Number of subgroups of $C_p^n\times C_q$: $2a(n,p)$. Number of subgroups of $C_p^n\rtimes C_q$: $a(n,p)+1+p^n$.

Thus the difference (which you expected to be non-negative) is $p^n+1-a(n,p)$.

For $n=1$ this is $p-2$, indeed non-negative. For $n=2$ this is $p^2-p-2=(p-2)(p+1)$, indeed non-negative. For $n=3$ this is $p^3+1-2(p+2)=p^3-2p-3$, indeed positive. For $n\ge 4$ this is $p^4+1-(p^4+p^3+2p^2+3p+5)=-(p^3+2p^2+3p+4)$, which is negative. For $n\ge 5$ it's certainly also negative, as $a(n,p)$ becomes huge and indeed them the direct product has much more many subgroups than the irreducible semidirect product.

So it's enough to find $p,q$ with an irreducible action of $C_q$ on $C_p^n$, $n\ge 4$. For instance for $q=5$ it's OK iff $X^4+X^3+X^2+X+1$ is irreducible modulo $p$. This is the case iff $q$ is not $1$ modulo $5$ (no nontrivial 5th root) and $X^2+X-1$ is irreducible, i.e. $5$ not a square mod $p$, which in turn holds iff $p$ is not square mod $5$, i.e. $p$ is $2$ or $3$ mod $5$ (i.e., $p=2$ or is $3$ or $7$ mod $10$).

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  • $\begingroup$ In summary: there's a duel between the $a(n,p)-2$ subgroups appearing only in the direct product, namely the $V\times C_q$, $V\subset C_p^n$ ($\neq 0,C_p^n$), and the $p^n$ conjugates of $C_q$ appearing in the nontrivial semidirect product (for in the direct case it's only a single one). The duel is won by the former for $n\ge 4$ and by the latter for $n\le 3$. $\endgroup$ Commented Nov 28 at 12:47
  • $\begingroup$ NB you can accept only one answer, you first accepted Sean's answer before mine was posted, you can accept it again, this additional answer didn't purport to steal this :) $\endgroup$ Commented Nov 28 at 12:49
  • $\begingroup$ My answer is much more Neanderthal (with a computer). $\endgroup$ Commented Nov 28 at 13:03
  • $\begingroup$ @YCor I accepted your answer, as it was more general in approach and it gives a concrete construction. Nevertheless, my sincere apologies to Professor Eberhard. $\endgroup$ Commented Nov 28 at 13:25
  • $\begingroup$ @YCor I've seen duals in math before, but this might be the first time I saw a duel in math! $\endgroup$ Commented Nov 28 at 19:34
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The smallest counterexample is $G = C_2 \times A_4 = H \rtimes K$ where $H = C_2^3$ and $K = C_3$. The group $G$ has $26$ subgroups but $H \times K = C_2^2 \times C_6$ has $32$.

There are $10$ counterexamples with $|G| \le 90$. There are a further $18$ counterexamples with $|G| = 96 = 2^5 \cdot 3$.

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