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Prove that a group of order 595 has a normal Sylow 17-subgroup.

The proof is as follows:

By Sylow, $n_{17} = 1$ or $35$. Assume $n_{17} = 35$. Then the union of the Sylow $17$-subgroups has $561$ elements. By Sylow, $n_5 = 1$. Thus, we may form a cyclic subgroup of order $85$ (from a previous theorem) But then there are $64$ elements of order $85$. This gives too many elements.

My question is: Where does this $64$ come from?

Mikasa
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Oliver G
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1 Answers1

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It comes from Euler's totient function, i.e., $$ \phi(85)=64. $$ Indeed, a cyclic group $C_n$ of order $n$ has exactly $\phi(n)$ generators, i.e., elements of order $n$.

The Phenotype
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Dietrich Burde
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  • Just a follow up question, how does this show that there's too many elements? It just seems to show that there's $64$ ways to represent that group of order $85$. – Oliver G Oct 07 '16 at 15:26
  • See the answers in the duplicate, i.e., count the elements. – Dietrich Burde Oct 07 '16 at 15:28
  • I can't quite see how the answers in the duplicate question answer the question I just asked. – Oliver G Oct 07 '16 at 15:35
  • Give it a try. If you do the calculations from the duplicate, you will immediately see what to do with $35(17-1)$ and $64$. – Dietrich Burde Oct 07 '16 at 15:37
  • I'm still don't understand. There's $64$ different elements that can be used to represent that specific subgroup of order $85$. How does the fact that there are $64$ different generators for that subgroup imply that there are too many elements? and what does $35(17-1) + 64$ represent? – Oliver G Oct 07 '16 at 15:44
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    @OliverG We've proven that there are at least $64$ elements of order $85$ and at least $560$ elements of order $17$ and one from order $1$ But as every order is unique to an element we have that there are at least $560+1+64=625$ elements in $G$. But $G$ is a group of order $595$, which is the number of elements in $G$. Obviously $625>595$ – Stefan4024 Oct 07 '16 at 16:03