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I have read a bunch of answers around the web, but they perform a jump which i can't follow.

I have determined that there must be $12*27=324$ elements of order 13 in G, but when i try to count the amount of elements in G of order 3 i run into some problems, i don't get the contradiction i was hoping for.

Some answers i found on the web, just says that since there is $3^3=27$ elements left in G, these must amount to a unique and therefore normal subgroup in G. But why is it so?

My proof so far:

Since $|G|=351=3^3*13$ Sylow theorem force $n_3$ to be $1$ or $13$, and $n_{13}$ to be $1$ or $27$. Show that neither $n_3$ nor $n_{13}$ is $1$, for contradiction. By lagrange theorem the order of the elements in the Sylow 13-subgroup must be $1$ or $13$, the only element with order $1$ is the $e$, which means there is $12$ elements, in each of the $27$ Sylow 13-subgroups, of order 13. Which means here are $12*27=324$ elements of order 13 in $G$. There remains $351-324=27=3^3$ elements in $G$, which must amount to a Sylow 3-subgroup, because?

I think i really need some Explain it like im 5 type of stuff here.

SuddenlyNotHorrific
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  • If the remaining 27 elements formed a normal subgroup, the group wouldn't be simple... – JustAskin Mar 11 '14 at 19:01
  • I edited the title, im sorry for the confusion. – SuddenlyNotHorrific Mar 11 '14 at 19:03
  • Have you learned about the 3 Sylow theorems? – user133458 Mar 11 '14 at 19:10
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    as @user133458 is saying, one of sylow's theorems guarentees the existence of a sylow p-subgroup. So you've assumed the sylow 13 subgroup was not unique, then showed there needed to be 27 of them, which left "room" for precisely 1 sylow 3 subgroup, hence its unique, hence it's normal. Then you can do it the other way round if you like to show if the sylow 3 isn't normal then the sylow 13 is. – snulty Mar 24 '14 at 03:58

1 Answers1

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If $n_3=1$, we are done. Suppose not, so there are 13 Sylows for 3. If there is one 13 Sylow there is nothing to prove, so suppose also there are 27 Sylows for 13. If $P$, $Q$ are two 13-subgroups then they are either disjoint or equal for every nontrivial element in them is a generator, since 13 is prime. This justifies the statment there are $27\cdot 12$ elements of order 13. This leaves 27 elements in $G$ not of order 13. In particular there are at most 27 elements with order a power of 3, contradicting the fact there are 13 distinct Sylow 3-subgroups.

Pedro
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  • I can't believe that this was all it took. Thank you so much! – SuddenlyNotHorrific Mar 11 '14 at 19:11
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    @SuddenlyNotHorrific No problem. It would be interesting if you shared the proof method in your post, i.e. provided some context. – Pedro Mar 11 '14 at 19:13
  • i have included my proof in the original post. – SuddenlyNotHorrific Mar 11 '14 at 19:28
  • @PedroTamaroff Why must an element of the Sylow 3-subgroup have order a power of 3? – Sam Oct 28 '14 at 02:12
  • @user151852 Sylow groups are $p$-groups. – Pedro Oct 28 '14 at 02:30
  • @PedroTamaroff So any element $x$ of a $p-group$ must have order a power of $p$ because its order must divide the order of the $p-group$ because we must have that the order of the cyclic subgroup generated by $x$ divides the order of the $p-group$ by Lagrange. Is that the correct reasoning? – Sam Oct 28 '14 at 17:47