# Proof explanation on a group of order $595$ having a normal Sylow $17$-subgroup. [duplicate]

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 PerryOct 8 '16 at 4:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

## 1 Answer

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 $n$.

• 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 7 '16 at 15:26
• See the answers in the duplicate, i.e., count the elements. – Dietrich Burde Oct 7 '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 7 '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 7 '16 at 15:37
• @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 7 '16 at 16:03