3
$\begingroup$

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 Answers 1

6

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

  • 0
    This confuses me: http://answers.yahoo.com/question/index;_ylt=Ai2F_HhB4LnFPwGpns2SfKnty6IX;_ylv=3?qid=20080503224602AAZapT4 Why is it 4 and not 5 in this case?2012-08-26
  • 2
    The subgroup of order 5 has 4 elements of order 5 and the identity which is of order 1.2012-08-26
  • 1
    So $G$ has $6$ elements of order $7$?2012-08-26
  • 5
    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 $e$.2012-08-26
  • 3
    @joachim: Yes. $H==\{1,x,x^2,x^3,x^4,x^5,x^6\}$ 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 $|x^m|=n/d$.2012-08-26
  • 0
    If I assume $G = \mathbb{Z}_{28}$ is $<7> = \{0,7,14,21,-7,-14,-21\}$ ?2012-08-27
  • 0
    Nope, because $-7 =21$. In your $G$, the element $7$ has order $4$. The elements of order $7$ in $\mathbb{Z}_{28}$ are $4, 8, 12, 16, 20, 24$. In general, in $\mathbb{Z}_{n}$ the oder of $m$ is $\operatorname{ord}(m)=\frac{n}{\gcd (m,n)}$.2012-08-27