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Could someone prove the sentence given above in the title?

I know that Sylow theorems should be used here.

Let $N_{p}$ stands for number of $p$-Sylow subgroups in group $G$.

I tried to use following sentences:

  • $\lvert G \rvert = 30 =2 \cdot 3 \cdot 5$
  • $N_{p} \equiv 1 \pmod p$
  • $N_{p} \mid 30$

But i can't figure out.

Regards.

Andreas Caranti
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mkultra
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  • While it doesn't really impact your result in this case. One important thing to note is Sylow's theorem actually guarantees us that $N_p|30/p$. This reduces the number of cases you have to check, for example, 6 has 4 divisors, while 30 has 8. – Melody Jan 14 '19 at 10:33
  • You could also assume that there is more than one Sylow $p$-subgroup for each prime and count elements of order $p$ for each of the primes. This leads to a contradiction. – Tobias Kildetoft Jan 14 '19 at 13:36

1 Answers1

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$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$If you know the result

if the finite group $G$ has order $\Size{G} = 2 d$, with $d$ odd, then $G$ has a subgroup of order $d$,

then you are immediately in business.

Otherwise, check the possibilities for the number of Sylow $5$-subgroups, and consequently for the number of elements of order $5$. If $N_{5} > 1$, you will find that there won't be too many elements left.

Andreas Caranti
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  • As regards to the given result- I am not sure whether this going to help. A group is not simple when there exists non-trivial normal subgroup not just ordinary subgroup. – mkultra Jan 14 '19 at 11:14
  • @mkultra Yes, but the result tells you that in a group $G$ of order $2d, d$ odd, you have a subgroup of order $d$, so $[G:H] = (2d)/d = 2$ and every subgroup of index $2$ is normal, you can see a proof here https://math.stackexchange.com/questions/84632/subgroup-of-index-2-is-normal – C_M Jan 14 '19 at 11:23
  • I want to only ask for proof of the result mentioned by @Andreas Caranti. – mkultra Jan 14 '19 at 13:33
  • @mkultra there is a proof here. – Andreas Caranti Jan 14 '19 at 14:28