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Problem $43$ on PDF-Page $20$ of: https://blngcc.files.wordpress.com/2008/11/rmc2007.pdf

Let $G$ be a finite group and $p_1, p_2, \ldots, p_n$ be distinct prime divisors of $|G|$, such that for each $p_i$ there exists $x_i, y_i$ in $G$ with $$\operatorname{ord}\left(x_i\right)=\operatorname{ord}\left(y_i\right)=p_i$$ and $y_i$ is not a power of $x_i$. Prove that $$ |G| \geq 1+\prod_{i=1}^n\left(p_i^2-1\right). $$


Update: $S_4$ is a counterexample as $$ |S_4|=4!=24<25=1+(2^2-1)(3^2-1) $$ and all the conditions for $S_4$ are fulfilled.

I suggest a new version of the Problem:

Let $G$ be a finite group and $p_1, p_2, \ldots, p_n$ be distinct prime divisors of $|G|$, such that for each $p_i$ there exists $x_i, y_i$ in $G$ with $$\operatorname{ord}\left(x_i\right)=\operatorname{ord}\left(y_i\right)=p_i$$ and $y_i$ is not a power of $x_i$. Prove that $$ |G| \geq 1+\sum_{i=1}^n\left(p_i^2-1\right). $$

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    $\begingroup$ I think $A_4$ (or even $S_4$) with $p_1=2$ and $p_2=3$ is a counterexample. $\endgroup$
    – Derek Holt
    Aug 28, 2023 at 19:48
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    $\begingroup$ Perhaps the question should have $\sum_{i = 1}^n$ instead of $\Pi_{i = 1}^n$. $\endgroup$ Aug 29, 2023 at 4:50
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    $\begingroup$ Yes it should be a sum rather than a product. Then you have to prove that $G$ has at least $p_i+1$ subgroups of order $p_i$ for each prime $p_i$, which is a nice exercise. $\endgroup$
    – Derek Holt
    Aug 29, 2023 at 7:17
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    $\begingroup$ No actually, @DerekHolt. But with that additional assumption I'd come up to the following (wrong?) proof. Every element is of some order: $$|G|\ge1+\sum_{i=1}^nk_i \tag1$$ where $k_i:=\left|\{g\in G\mid o(g_i)=p_i\}\right|$. By assumption, for every $i$ there are at least two (distinct) subgroups of order $p_i$, namely $H_i:=\langle x_i\rangle$ and $K_i:=\langle y_i\rangle$. If either $H_i\unlhd G$ or $K_i\unlhd G$, then $H_iK_i\cong C_{p_i}\times C_{p_i}$, so there are at least $p_i^2-1$ elements of order $p_i$, namely: $$k_i\ge p_i^2-1$$ which replaced in $(1)$ gives the claim. $\endgroup$
    – Kan't
    Aug 30, 2023 at 7:47
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    $\begingroup$ Ok, I'll think some more about that. $\endgroup$
    – Kan't
    Aug 30, 2023 at 7:53

1 Answer 1

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Here is a sketch proof of the amended version of the problem with the product replaced by sum. That is $$|G| \ge 1 + \sum_{i=1}^n(p_i^2-1).$$ It is enough to prove that the for each $i$ the number of $p_i$-elements of $G$ is at least $p_i^2-1$. This is clear if a Sylow $p_i$-subgroup $P_i$ of $G$ has order at least $p_i^2$, so we just have to consider the case when $|P_i|=p_i$.

By the assumption that we have elements $x_i$ and $y_i$ of order $p_i$, where $x_i$ is not a power of $y_i$, $P_i$ cannot be the unique subgroup of $G$ of order $p$, so by Sylow's Theorem $G$ has at least $p_i+1$ Sylow $p_i$-subgroups, which together contain at least $(p_i+1)(p_i-1)=p_i^2-1$ elements of order $p_i$.

I think the only examples in which we get equality are $A_4$ and groups $C_p \times C_p$ for primes $p$, but I have not proved that.

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    $\begingroup$ Slightly different argument: By a generalization of Sylow's theorem, the number of subgroups of order $p_i$ is always $\equiv 1 \mod{p_i}$. So there are at least $p_i+1$ subgroups of order $p_i$ in $G$, giving at least $p_i^2-1$ elements of order $p_i$. $\endgroup$ Aug 31, 2023 at 2:40

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