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Let $G$ be a group of order $|G|=6545$. Show that $G$ has either a normal Sylow 5-subgroup or a normal Sylow 17-subgroup.

My attempt (supposedly wrong):

Given $|G|=6545=5\cdot7\cdot11\cdot17$, by Sylow's theorem, we have: $n_{5}=1\text{ mod }5\text{ and }n_{5}\mid7\cdot11\cdot17=1309\Rightarrow n_{5}=1\text{ or }11\\n_{7}=1\text{ mod }7\text{ and }n_{7}\mid5\cdot11\cdot17=935\Rightarrow n_{7}=1\text{ or }85\\n_{11}=1\text{ mod }11\text{ and }n_{11}\mid5\cdot7\cdot17=595\Rightarrow n_{11}=1\text{ or }595\\n_{17}=1\text{ mod }17\text{ and }n_{17}\mid5\cdot7\cdot11=385\Rightarrow n_{17}=1\text{ or }35$

Suppose $G$ has no normal Sylow 5-subgroup or normal Sylow 17-subgroup. Then, we have $n_{5}=11$ and $n_{17}=35$. and there exist $11\cdot(5-1)+35\cdot(17-1)=604$ non-identity elements of order $5$ and $17$. Consequently we have $6545-604=5941$ elements of order 7 or 11. But this is a impossible because the only possible outcomes of $n_{7}$ and $n_{11}$ are:$1\cdot(7-1)+1\cdot(11-1)+1=17\neq5941\\1\cdot(7-1)+595\cdot(11-1)+1=5956\neq5941\\85\cdot(7-1)+1\cdot(11-1)+1=521\neq5941\\85(7-1)+595\cdot(11-1)+1=6461\neq5941$

Thus, we must have $n_{5}=1$ or $n_{17}=1$. In other words, $G$ has either a normal Sylow 5-subgroup or a normal Sylow 17-subgroup.

I know my solution is completely wrong and I'm not supposed to solve it this way. Can someone help me?

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    $\begingroup$ You can't say "we have $5941$ elements of order $7$ or $11$". First, there is an element of order $1$ and there may also be elements of order $5 \cdot 7$ for instance. $\endgroup$ – v_lentin May 9 '18 at 23:04
  • $\begingroup$ Assume $n_5=11$ and look at the (transitive) action of $G$ on the set of its $11$ Sylow $5$-subgroups. What do you know about the kernel of this action? $\endgroup$ – j.p. May 10 '18 at 7:36
  • $\begingroup$ @CMonsour let’s not assume. first of all this is not for a class. And I did not downvote your answer. $\endgroup$ – Luis Defendi May 10 '18 at 13:23
  • $\begingroup$ @CMonsour just because it is on my question, doen’t mean I’m the only one that can vote. I didn’t downvote your answer. What js your problem. $\endgroup$ – Luis Defendi May 10 '18 at 13:24
  • $\begingroup$ @CMonsour If you want an evidence that I didn’t downvote you. I can downvote you right now. Does that make you happier? $\endgroup$ – Luis Defendi May 10 '18 at 13:25
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Here is one way to proceed.

  • First case: $n_{5} = 1$.
    Then $G$ has a normal Sylow $5$-subgroup.
  • Second case: $n_{5} = 11$.
    Let's consider the action of $G$ on the set of its Sylow $5$-subgroups by conjugation i.e the morphism $$\pi : \begin{array}[t]{rcl} G & \longrightarrow & \mathfrak{S}_{\left\{ \text{Sylow } 5\text{-subgroups of } G \right\}}\\ g & \longmapsto & \left(S \mapsto g S g^{-1}\right) \end{array}$$
    Then $\frac{|G|}{|\text{Ker}(\pi)|} = |\text{Im}(\pi)| \geq 11$ since this action is transitive - all the Sylow $5$-subgroups are conjugate to each other - and divides both $|G| = 5 \cdot 7 \cdot 11 \cdot 17$ and $|\mathfrak{S}_{\text{Sylow } 5\text{-subgroups}}| = 11!$
    Therefore, $$\frac{|G|}{|\text{Ker}(\pi)|} = 11, \ 5 \cdot 7, \ 5 \cdot 11, \ 7 \cdot 11 \text{ or } 5 \cdot 7 \cdot 11$$ and hence, $$|\text{Ker}(\pi)| = 17, \ 5 \cdot 17, \ 7 \cdot 17, \ 11 \cdot 17 \text{ or } 5 \cdot 7 \cdot 17$$
    Furthermore, $|\text{Ker}(\pi)| \ne 5 \cdot 17 \text{ and } 5 \cdot 7 \cdot 17$.
    Indeed, let's suppose that $|\text{Ker}(\pi)| = 5 \cdot 17 \text{ or } 5 \cdot 7 \cdot 17$.
    Then, since $v_{5}(|G|) = v_{5}(|\text{Ker}(\pi)|)$ and $\text{Ker}(\pi) \vartriangleleft G$, the Sylow $5$-subgroups of $G$ are exactly the Sylow $5$-subgroups of $\text{Ker}(\pi)$.
    Hence, $n_{5} = 11 | 7 \cdot 17$. Contradiction.
    Thus, $$|\text{Ker}(\pi)| = 17, \ 7 \cdot 17 \text{ or } \ 11 \cdot 17 $$
    Now, since $v_{17}(|G|) = v_{17}(|\text{Ker}(\pi)|)$ and $\text{Ker}(\pi) \vartriangleleft G$, the Sylow $17$-subgroups of $G$ are exactly the Sylow $17$-subgroups of $\text{Ker}(\pi)$.
    Hence, $n_{17} \equiv 1 \pmod{17}$ and $n_{17} | 7 \cdot 11$ which implies that $n_{17} = 1$.
    Thus, $G$ has a normal Sylow $17$-subgroup.
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  • $\begingroup$ What does the notation $\nu_5$ mean? $\endgroup$ – Luis Defendi May 10 '18 at 2:26
  • $\begingroup$ If $n$ is an integer $\geq 2$ and $p$ is a prime number, $v_{p}(n)$ denotes the greatest integer $\alpha$ such that $p^{\alpha}$ divides $n$. $\endgroup$ – v_lentin May 10 '18 at 10:33
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The 17-Sylow subgroup of a group of this order is ALWAYS normal.

Simple solution: First, a group of this order is solvable. (You don't need the odd order theorem...groups of square-free order can be shown to be solvable by the Burnside transfer theorem, and by other means..there is a non-transfer proof in Marshall Hall's Theory of Groups that a square-free order group is metacyclic...i.e., a a normal cyclic subgroup with cyclic factor group.) For solvable groups, Philip Hall's theorem (see, e.g., Th 9.3.1 in Marshall Hall's book) says that the number of Sylow-17 subgroups is a product of factors each of which is congruent to $1\pmod{17}$ and divides a chief factor. Since the chief factors of a (necessarily solvable) square-free group are the primes dividing it, the number of 17-Sylow subgroups must be a product of ones..i.e., 1.

Thus, in fact a much more general result holds: ANY group of square-free order has a normal Sylow subgroup corresponding to its largest prime.

In fact, a group of order 6545, by the above result, must have normal subgroups of orders 7, 11, and 17, and thus has a cyclic normal subgroup of index 5, which has automorphism group a direct product of cyclic groups of orders 6, 10, and 16, and that has a unique subgroup of order 5. Thus, a Sylow-5 element has two ways to act on the index 5 subgroup...trivially or faithfully. This means that, up to isomorphism, there are only two groups of order 6545--the cyclic group, and a semidirect product with a 5-element acting non-trivially on an index 5 normal cyclic subgroup. Note that another way to characterize the non-abelian group of order 6545 is as the direct product of a cyclic group of order 119 with a Frobenius group of order 55.

I should add that if you have a reference that does not contain the counting statement of Philip Hall's theorem but only the other parts of it (e.g., Isaacs's book), I posted a proof of the counting statement as it applies to Sylow subgroups here: Sylow Counting Generalization of Hall Theorem . Note that even though this statement for Sylow subgroups applies to all finite groups you still need to show independently that the group is solvable, since that's how you know that the chief factors are all 17 or less.

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