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Let's say we have a group G of order ninety and want to see whether the case $n_5=1$, $n_2=5$ and $n_3=10$ exists (where $n_p$ is the number of P-sylow subgroups).

Proof

This cannot happen, because, since $P_5 \vartriangleleft G$; $P_5P_3$ must be a subgroup of order 45 (which has index 2 and is normal) but there can't be a subgroup of order 45 unless $P_3 \vartriangleleft G$.

I don't really understand why $P_3$ would need to be normal if a group of order 45 exists...

Thanks in advance

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    $\begingroup$ For fixed $p$, every Sylow $p$-subgroup is conjugate, and every conjugate of a given Sylow $p$-subgroup is another Sylow $p$-subgroup. Now, a subgroup has only one conjugate (itself) if and only if it is a normal subgroup. By hypothesis, $n_5=1$, so $P_5\triangleleft G$. $\endgroup$ – anon May 4 '13 at 16:15
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Consider the number $m_3$ of Sylow $3$-subgroups in the subgroup $N$ of order $45$. By Sylow's theorems, $m_3$ must divide $5$, and be congruent to $1$ modulo $3$, so $m_3 = 1$. So the unique Sylow $3$-subgroup $T$ of $N$ is characteristic in $N$, which is normal in $G$, and thus $T$ is normal in $G$. So $n_3 = 1$, contradicting the assumption that $n_3 = 10$.

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  • $\begingroup$ simple and straightforward ++. $\endgroup$ – mrs May 5 '13 at 19:07
  • $\begingroup$ @BabakS. Thanks! $\endgroup$ – Andreas Caranti May 5 '13 at 19:24

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