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A proof that a group of order $180$ is not simple is given here.

However, the proof uses Burnside $p$-complement theorem. If you know a proof without Burnside $p$-complement theorem, please let me know. I understand Burnside theorem.

Here is a summary of what I understand

a. $n_5=1,6$ or $36$.

b. $n_5 \neq 1$ is obvious.

c. $n_5=6$ is a contradiction by embedding $G$ into $A_6$.

Therefore, we only need to consider the case $n_5=36$.

Akasa
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  • Is the answer in the link based on Bunrnside $p$-complement theorem? – Kan't Jun 10 '24 at 13:27
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    @Kan't Yes, it is. Because it assumes $n_5=6$, picking up from where the asker stopped. And the asker excluded the possibility $n_5=36$ using Burnside. – Jyrki Lahtonen Jun 11 '24 at 04:01

1 Answers1

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I think a two-pronged attack using both Sylow $5$s and Sylow $3$s will get the job done.

As you observed, we can assume that $n_5=36$ for otherwise we run into a contradiction with the simplicity of $A_6$. Therefore there are $36(5-1)=144$ elements of order five.

Similarly, $n_3=4$ is out of the reckoning, so we can assume that $n_3=10$. All the Sylow $3$-subgroups have $9=3^2$ elements, so they are abelian. Assume that $x\neq1$ is in the intersection of two Sylow $3$-subgroups $P$ and $P'$. Then $x$ is centralized by both $P$ and $P'$, so $H=C_G(x)$ has more than nine elements, and more than one Sylow $3$-subgroup. Hence $n_3(H)\in\{4,10\}$, so $|H|\ge 36$. Of course, if $H=G$, then $x\in Z(G)$ and we are done. In the other cases $[G:H]\in\{2,5\}$, and again the action of $G$ on the set of cosets of $H$ gives a homomorphism from $G$ with a non-trivial kernel.

The remaining possibility is that any pair of Sylow $3$-subgroups has a trivial intersection. In that case there are $10(9-1)=80$ elements of orders $3$ or $9$. But, together with the elements of order five that's too many.

Jyrki Lahtonen
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  • My brain is not very agile with this kind of arguments, so something simpler may be out there. – Jyrki Lahtonen Jun 10 '24 at 06:43
  • Thank you for your thoughtful answers. – Akasa Jun 10 '24 at 07:51
  • This is probably best possible avoiding Burneside Transfer Theorem – Derek Holt Jun 10 '24 at 10:14
  • Jyrki, I saw you commented on my question about a simple unsolvable quintic with exactly one real root, but now those comments seem to have been deleted. Is my question clear to you, or do you still not get what I'm asking for? Thanks! – user21820 Jun 27 '24 at 11:40