# Romanian Math Competition (2007): Group Cardinality Inequality $|G| \geq 1+\prod_{i=1}^n\left(p_i^2-1\right)$

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 $$|G| \geq 1+\sum_{i=1}^n\left(p_i^2-1\right).$$

• I think $A_4$ (or even $S_4$) with $p_1=2$ and $p_2=3$ is a counterexample. Aug 28, 2023 at 19:48
• Perhaps the question should have $\sum_{i = 1}^n$ instead of $\Pi_{i = 1}^n$. Aug 29, 2023 at 4:50
• 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. Aug 29, 2023 at 7:17
• 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. Aug 30, 2023 at 7:47
• Ok, I'll think some more about that. Aug 30, 2023 at 7:53

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.
• 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 $p_i$. Aug 31, 2023 at 2:40