5

As I said in the title, the question requires $G$ is not simple with the $\vert G \vert =160$. But All the the solution I've looked, it only considered the case for the Sylow 2 group case. So I tried different ways to show not simple taking the Sylow 5 group case.

Here is my solution .

$sol)$ $160 = 2^5 \bullet 5 $, So the number of the Sylow 5 group, $n_5 \in \{1,16\}$

First case) $n_5 = 1 $, Then $P \lhd G$ for the Sylow 5 group, $P$ (Not simple)

Second case) Suppose $n_5 =16$. The $N(H)$ is a normalizer of the $H$

Taking two different Sylow 5 group, $P_1$ and $P_2$, Hence $\vert P_1 \cap P_2 \vert =1$

Then $P_1 \cap P_2 \lhd P_1$ and $P_1 \cap P_2 \lhd P_2$ $\Rightarrow$ $P_1 \cup P_2 \subset N(P_1 \cap P_2)$

Plus since $N(P_1 \cap P_2) \leq G$, $N(P_1 \cap P_2) \vert \vert G \vert$ $\Rightarrow$ $\vert N(P_1 \cap P_2) \vert =5$ Since $P_i \leq N(P_1 \cap P_2)$

So the conclusion is $\vert N(P_1 \cap P_2) \vert = \vert P_1 \vert = \vert P_2 \vert $, Therefore $P_1 = P_2 = N(P_1 \cap P_2)$

It is a contradict with the there are different 16 Sylow 5 groups.

At least I thought, It looks fine to me. But I can't sure my solution is exact or not. Is my solution right? Any advice and answer always welcome. Thanks.

Bernard
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se-hyuck yang
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    Why not show there is a single $2$-Sylow? It is simpler. – Bernard Jul 07 '20 at 11:20
  • @Bernard, I just tried the 5-sylow case. Because all the solution solved it by 2 sylow as I mentioned in the post. – se-hyuck yang Jul 07 '20 at 11:33
  • In this case, it makes sense to use the prime $2$ , but in other situations , we might need such an argument. So , I think the question whether this is valid is useful. (+1) – Peter Jul 07 '20 at 11:46

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I don't think your proof works. Once we know that $P_1 \cap P_2$ is trivial, we know that $N(P_1 \cap P_2) = G$. Nothing in your proof is inconsistent with the normalizer being $G$ until you somehow deduce that $\vert N(P_1 \cap P_2) \vert =5$. In any case, you can't prove that every group of order $160$ has a normal Sylow $5$-subgroup, because it isn't true. Here's a counterexample:

Let $N = C_2 \times C_2 \times C_2 \times C_2 \times C_2$, and let $\alpha$ be the automorphism of $N$ defined by $\alpha((a_1,a_2,a_3,a_4,a_5)) = (a_5,a_1,a_2,a_3,a_4)$. The order of $\alpha$ is $5$, so we can use $\alpha$ to construct a semidirect product $G = N \rtimes C_5$. Then $G$ is a group of order $160$ and $G$ does not have a normal Sylow $5$-subgroup.

Zoe H
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