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Existence of a $\mathbb{Z} _{15}$ in a group $G$ where $|G|=3 \cdot 5 \cdot 17$.

This is my proof:

we know (from a previous point of the exercise) that $\mathbb{Z}_{17}$ is a characteristic subgroup of $G$. For Cauchy, we have that $\exists \mathbb{Z}_3,\ \mathbb{Z}_5 <G$. If we demonstrate that $\mathbb{Z}_3$ or $\mathbb{Z}_5$ is normal in $G$, then we have that $\mathbb{Z}_3\mathbb{Z}_5$ (or $\mathbb{Z}_5\mathbb{Z}_3$) is a subgroup of $G$ of order 15, therefore is isomorphic to $\mathbb{Z}_{15}$.

For Sylow, we have that $n_5 \equiv 1 \pmod{5}$ and $n_5 \mid 51$ where $n_5$ is the number of $5$-Sylow in $G$, so $n_5=1,51$.

If $n_5=1$ then the only $5$-Sylow in $G$ is normal.

If $n_5=51$ then there are $51 \cdot 4$ elements of order $5$ in $G$, and of the remaining $51$ elements there are $16$ of order $17$ (from the only $\mathbb{Z}_{17}$) and one of order $1$, that is the identity. So we have $34$ elements left.

At this point, I thought about two different conclusions, but I am not really sure which one is the correct one:

  1. All the $34$ elements have to be of order $3$, therefore there are $17$ $3$-Sylow, but this is impossible because $17 \not\equiv 1 \pmod{3}$. So we can conclude that $n_5=1$.

  2. We have $34$ elements, so we can have at most $17$ $3$-Sylow (in other words, $n_3\le 17$), and this means that $n_3$ can only be $1$. So if $n_5 \neq 1$ then $n_3=1$.

azimut
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Luigi Traino
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    Both are essentially correct: There aren’t enough elements left to have more than one $3$-Sylow subgroup when there is more than one $5$-Sylow, so there is either a normal subgroup of order $5$, or a normal subgroup of order $3$. And from there the result will follow. I will note, however, that in general it is not true that for $HK$ to be a subgroup you need either $H$ or $K$ to be normal; that is sufficient, but not necessary. You can have $HK=KH$ as sets (and hence $HK$ a subgroup), with neither $H$ nor $K$ normal. It’s not even necessary that one normalize the other. – Arturo Magidin Jan 01 '21 at 20:49
  • Try using the normalizer centralizer theorem applied to your Sylow 17 subgroup. – Steve D Jan 01 '21 at 20:51
  • What bothers me is that these two arguments bring to two different conclusions. In the first case, you proved that $n_5=1$, while in the second case you proved that $n_5$ can be $1$, and if it's not, then $n_3=1$. I mean, these two conclusion cannot both be true. – Luigi Traino Jan 01 '21 at 21:55
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    The last sentence in 1 is indeed not correct: the contradiction does not come from “$n_5\neq 1$”, it comes from “all remaining elements have order $3$; it just means that the remaining elements cannot all be in $3$-Sylow subgroups; that’s why I said “essentially correct”: you correctly conclude that once you remove the elements of orders $17$, $5$, and $1$, not all remaining elements can have order $3$, so not all remaining elements are in $3$-Sylow subgroups, so there aren’t enough elements to have more than one $3$-Sylow subgroup. – Arturo Magidin Jan 01 '21 at 22:56
  • This question is covered by https://math.stackexchange.com/q/255441/61691, where it is shown that every group of order 255 is abelian (and hence cyclic). – azimut Aug 22 '22 at 08:58

2 Answers2

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$\textbf{Hint}:$ let's take a look at the number of groups in $Syl_{17}(G)$ and $Syl_3(G)$.
We know that $n_{17}\equiv1\text{ (mod 17)}\implies n_{17}\in \{1\not3,\not5,\not{15}\}\implies \exists!P_{17}\trianglelefteq G$.
While $n_3\equiv 1\text{ (mod 3)}\implies n_3\in\{1,\not5,\not17,85\}.$
If we now consider the action of conjugation of $G$ on its Sylow's groups we have that $$|Syl_{17}(G)|=|G:N_G(P_{17})|\implies|N_G(P_{17})|=|G|.$$ Furthermore $$\dfrac{|N_G(P_{17})|}{|C_G(P_{17})|}\bigg||\operatorname{Aut}(P_{17})|=\varphi(17)=16\implies\dfrac{3\cdot 5\cdot 17}{|C_G(P_{17})|}\big |2^4\iff|C_G(P_{17})|=|G|.$$ $P_{17}$ is abelian, so $P_{17}\le N_G(P_3)\implies17\big||N_G(P_{3})|$. We also have $\{1,85\}\ni n_3=|G:N_G(P_3)|$, so $n_3=1$. You can do something similar for $P_5$.

Vajra
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As also indicated in the comments, there is a fast argument based on counting elements of order $3$ and $5$. It has also been used as a first step here.

By Sylow, $n_3\in \{1,85\}$ and $n_5\in\{1,51\}$.

Assume that $n_3 = 85$ and in the same time $n_5 = 51$. Then $G$ contains $85\cdot 2 = 170$ elements of order $3$ and $51 \cdot 4 = 204$ elements of order $5$, which is not possible since $G$ contains only $255$ elements.

So $n_3 = 1$ or $n_5 = 1$. Let $P_3$ be a $3$-Sylow subgroup and $P_5$ a $5$-Sylow subgroup. Then at least one of the groups $P_3$ and $P_5$ is normal in $G$, implying that $P_3 P_5$ is a subgroup of $G$. Since $P_3$ and $P_5$ have trivial intersection, $\#(P_3 P_5) = 3\cdot 5 = 15$. Now by the classification of $pq$-groups and $5\not\equiv 1\bmod 3$, the subgroup $P_3 P_5$ is cyclic of order $15$.

(In fact, much more can be shown, namely that every group of order $3\cdot 5\cdot 17 = 255$ is abelian and hence cyclic (by the classification of finite abelian groups, since 255 is square-free). Therefore, there is a unique subgroup of order $15$, which is cyclic.)

azimut
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