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Take a group $G$ of order $\lvert G \rvert = 68 = 2^2\cdot17$. Let $n_2$ be the number of $2$-Sylow Subgroups of $G$ and $n_{17}$ be the number of $17$-Sylow Subgroups of $G$. Since $\lvert G \rvert = 2^2\cdot17$, we have

\begin{align} n_2\vert17\ \ \ &\text{and}\ \ \ n_2 \equiv 1\mod{2}\\ n_{17}\vert 4\ \ \ &\text{and}\ \ \ n_{17}\equiv 1 \mod{17} \end{align}

Since the only $n_{17}$ that divides $4$ which is also congruent to $1$ modulo $17$ is $1$, we must have that $n_{17} = 1$, while for $n_2$ we have that $n_2 = 1, 17$.

I'm confused, though, that $n_2$ could be equal to $17$, because $G$ must have a subgroup of order $17$, but if $n_2 = 17$ then we have $17$ subgroups of $G$ each of order $4$, which leaves no room for a subgroup of order $17.$

Have I made a silly mistake/is my understanding off?

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$17$ subgroups of order $4$ at least all intersect in the identity. So that means they consist together of at most $$1+17(4-1)=52$$ elements. Now we've already accounted for the identity, so a subgroup of order $17$ only adds $16$ more elements, which adds up to $68$.

In general, the subgroups of order $4$ could share more elements than just the identity, so their union would have even fewer than $52$ elements.

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  • $\begingroup$ Of course, I was counting the identity 17 times. Thanks. $\endgroup$ – ÍgjøgnumMeg Jun 18 '17 at 15:25
  • $\begingroup$ @Ig No problem. $\endgroup$ – Matt Samuel Jun 18 '17 at 15:26

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