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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 $7$?

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

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    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