This question already has an answer here:

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?


marked as duplicate by Dietrich Burde, Willie Wong, Parcly Taxel, apnorton, JonMark Perry Oct 8 '16 at 4:51

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  • $\begingroup$ $ \varphi(85) = 64 $. $\endgroup$ – Starfall Oct 7 '16 at 15:20
  • $\begingroup$ What's $\phi(85)$ represent? $\endgroup$ – Oliver G Oct 7 '16 at 15:21
  • $\begingroup$ See Dietrich Burde's answer. $\endgroup$ – Starfall Oct 7 '16 at 15:21

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 $n$.

  • $\begingroup$ 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 $85$. $\endgroup$ – Oliver G Oct 7 '16 at 15:26
  • $\begingroup$ See the answers in the duplicate, i.e., count the elements. $\endgroup$ – Dietrich Burde Oct 7 '16 at 15:28
  • $\begingroup$ I can't quite see how the answers in the duplicate question answer the question I just asked. $\endgroup$ – Oliver G Oct 7 '16 at 15:35
  • $\begingroup$ Give it a try. If you do the calculations from the duplicate, you will immediately see what to do with $35(17-1)$ and $64$. $\endgroup$ – Dietrich Burde Oct 7 '16 at 15:37
  • 1
    $\begingroup$ @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 $625>595$ $\endgroup$ – Stefan4024 Oct 7 '16 at 16:03

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