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Let $G$ be a group of order $48$. By the $1$st Sylow theorem $G$ has a Sylow $2$-subgroup and a Sylow $3$-subgroup. Suppose none of these are normal. Determine the number of Sylow $2$-subgroups and Sylow $3$-subgroups that $G$ can have. Justify.

Let $G$ be a group of order $48=3 \cdot 2^4$. The number of Sylow $2$-subgroups $n_2$ divides $24$ and has the form $n_2=2k+1$ by the Sylow Theorems. Therefore $n_2=1,3$. However since we want non-normal subgroups so $n_2=3$.

The number of Sylow $3$-subgroups $n_3$ divides $16$ and has the form $n_3=3k+1$ by the Sylow Theorems. Therefore $n_3=1,4$. However since we want non-normal subgroups so $n_3=4$.

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    $\begingroup$ ...and why not $\;n_3=16\;$ ? $\endgroup$ – Timbuc Dec 4 '14 at 4:07
  • $\begingroup$ I forgot about that one. So other than that it is ok? $\endgroup$ – Username Unknown Dec 4 '14 at 5:02
  • $\begingroup$ Yup...and it can't really be $\;16\;$ , either. Count elements. :) $\endgroup$ – Timbuc Dec 4 '14 at 5:10
  • $\begingroup$ How would I determine how many elements are in each subgoup? For example, is it 48/16=3? $\endgroup$ – Username Unknown Dec 4 '14 at 5:16
  • $\begingroup$ I meant to count elements in the whole group. $\endgroup$ – Timbuc Dec 4 '14 at 10:36
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Assume that $n_3 = 16$. If $P_i, P_j \in Syl_3(G)$ and $P_i \neq P_j$, then $P_i \cap P_j = 1$. Each Sylow $3$-subgroup has $3$ elements, the identity and two others. The intersections only contain the identity, and hence from each Sylow $3$-subgroup you can count $2$ new elements.

Now $n_2 = 3$. If $Q_i, Q_j \in Syl_2(G)$ and $Q_i \neq Q_j$, then $$ |Q_iQ_j| = \frac{|Q_i||Q_j|}{|Q_i \cap Q_j|} = \frac{16 \cdot 16}{|Q_i \cap Q_j|} \leq 48. $$ Now $|Q_i \cap Q_j| \ | \ |Q_i| = 16$, thus $|Q_i \cap Q_j| = 8$ or $16$. As $Q_i \neq Q_j$, then $Q_i \cap Q_j \neq Q_i$ and you can deduce that $|Q_i \cap Q_j| = 8$. In the worst case (least amount of distinct elements), you get $3 \cdot 8$ new elements from the $3$ Sylow $2$-subgroups.

In total you get $16 \cdot 2 + 3 \cdot 8 = 56 > 48$ elements, which is a contradiction. Thus $n_3 = 16$ and $n_2 = 3$ is not possible.

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  • $\begingroup$ How do you know that there is 2 new elements? Also how did you deduce $|Q_i \cap Q_j|=8$? $\endgroup$ – Username Unknown Dec 4 '14 at 18:58
  • $\begingroup$ I edited my answer to contain more details. $\endgroup$ – Leppala Dec 5 '14 at 8:03
  • $\begingroup$ Got it. It makes so much more sense now thank you for your help $\endgroup$ – Username Unknown Dec 5 '14 at 15:28
  • $\begingroup$ I have another question how do we now that $n_3=4$ is possible as well? $\endgroup$ – Username Unknown Dec 6 '14 at 21:21

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