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 $7$?


1 Answer 1


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

  • $\begingroup$ This confuses me: answers.yahoo.com/question/… Why is it 4 and not 5 in this case? $\endgroup$
    – Joachim
    Aug 26, 2012 at 13:48
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    $\begingroup$ The subgroup of order 5 has 4 elements of order 5 and the identity which is of order 1. $\endgroup$
    – neelp
    Aug 26, 2012 at 13:51
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    $\begingroup$ So $G$ has $6$ elements of order $7$? $\endgroup$
    – Joachim
    Aug 26, 2012 at 13:53
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    $\begingroup$ 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 $e$. $\endgroup$
    – N. S.
    Aug 26, 2012 at 13:56
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    $\begingroup$ @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 $|x^m|=n/d$. $\endgroup$
    – Mikasa
    Aug 26, 2012 at 14:03

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