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Let $G$ be a finite group, and let $P_1, ..., P_r$ be its nontrivial Sylow subgroups. Assume all $P_i$ are normal in $G$. I need to prove that $(P_1...P_{r-1}) \cap P_r = \{ e \}$.

I know that $P_1...P_{r-1}$ is a normal subgroup of $G$ (since all $P_i$ are normal). But I don't know what is its order $|P_1...P_{r-1}|$, though, if it's important (could use Lagrange's theorem). I also know the case for $r = 3$, by Second Isomorphism Theorem we have $$|P_1P_2| = \frac{|P_1||P_2|}{|P_1 \cap P_2|} = |P_1||P_2| = p^{m_1}_1p^{m_2}_2\,.$$ Since $\gcd(|P_1P_2|,|P_3|) = 1$, then $P_1P_2 \cap P_3 = \{ e \}$.

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If each Sylow subgroup is normal, then there is a unique Sylow subgroup for each prime dividing the order of the group. Every element of $P_r$ will have order a power of some prime $p_r$, while each other $P_i$ contains only elements of order relatively prime to $p_r$.

More formally, for $x \in (P_1\dotsb P_{r-1})\cap P_r$, we must have that $x$ has order $p_r^k$ because it is in $P_r$ and also have order $p_1^{k_1}\dotsb p_{r-1}^{k_{r-1}}$ because it is in $P_1\dotsb P_{r-1}$ where $\gcd(p_r,p_1\dotsb p_{r-1}) = 1$. This can only be the case where $k = k_1 = \dotsb = k_{r-1} = 0$, so $x$ has order $0$ and must be the identity.

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  • $\begingroup$ So, let $g_i \in P_i$. Assume $g_1...g_{r-1} = g_r$. $|g_r|$ divides $|P_r| = p^{m_r}_r$, so $|g_r| = p^n_r n \leq m_r$. On the other hand, $|g_1...g_{r-1}|$ divides $lcm(|g_1|, ..., |g_{r-1}|) = {p^{n_1}}_1...p^{n_{r-1}}_{r-1}$. Since $\gcd(p^n_r, p^n_1_1...p^{n_{r-1}}_{r-1}) = 1, |g_r| = |g_1...g_{r-1}| = 1$, and $g = g_1...g_{r-1} = e$. Right? $\endgroup$ Commented Aug 21, 2016 at 1:15
  • $\begingroup$ @Jxt921, yeah, I think it could be worded better, but you have the correct idea now. $\endgroup$ Commented Aug 21, 2016 at 2:00

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