Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a group $G$ with order $28 = 2^2 \cdot 7$. Sylow-Theory implies that there is a exactly one $7$-Sylow-Subgroup of order $7$ in $G$, and $1$ or $7$; $2$-Sylow-Subgroups.

Where to go from here concerning the number of elements of order $7$?

share|improve this question

1 Answer 1

up vote 6 down vote accepted

If $x$ is an element of order $7$ then the subgroup $\langle x\rangle$ generated by $x$ has order $7$...

share|improve this answer
    
This confuses me: answers.yahoo.com/question/… Why is it 4 and not 5 in this case? –  joachim Aug 26 '12 at 13:48
2  
The subgroup of order 5 has 4 elements of order 5 and the identity which is of order 1. –  neelp Aug 26 '12 at 13:51
1  
So $G$ has $6$ elements of order $7$? –  joachim Aug 26 '12 at 13:53
4  
Denote $H$ the only subgroup of order 7 in $G$. We just proved that if $x$ is an element of order $7$, then $x \in H$. Now you can use (prove if you don't know it) that $H$ is cyclic, and has six elements of order $7$. The seventh element of $H$ is $e$. –  N. S. Aug 26 '12 at 13:56
3  
@joachim: Yes. $H=<x>=\{1,x,x^2,x^3,x^4,x^5,x^6\}$ where $x\in G$ has order $7$. Note that in any group, say $G$, if $x\in G$, $|x|=n$ and $(m,n)=d$ then $|x^m|=n/d$. –  Babak S. Aug 26 '12 at 14:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.