Abstract: If $G$ is a group of order 10 with only one element of order 2, then $G$ must be cyclic. Abstract Algebra:
If $G$ is a group of order 10 with only one element of order 2, then $G$ must be cyclic.
Thank you. 
 A: There will be only one Sylow  $5$ subgroup $\Rightarrow$ there are $4$ elements of order $5$ also there is $1$ element of order $2$ and $1$ element of order $1$ i.e. identity. Thus remaining $4$ elements must be of order $10$ because order of an element divides the order of the group. And element of order $10 \Rightarrow$ G is cyclic. 
A: Sylow theory is not even necesary for this problem.
Let $x$ be the unique element of order $2$.
Let $y$ be an element not equal to $x$ or $e$. By Lagrange's Theorem, $y$ can have order $1$, $2$, $5$, or $10$. Since $x$ is the only element of $G$ with order two and $y \neq e$, we know that $y$ does not have order $1$ or $2$. Furthermore, if $y$ had order $10$, then $G$ would be cyclic and we would be done. Hence we only need to consider the case where $y$ has order $5$.
Clearly all powers of $y$ must also have order $5$, so let us enumerate some of the elements of $G$:
$$ \{ e,x,y,y^2,y^3,y^4 \} \subset G$$
Now let $z$ be an element in $G$ not equal to any of the elements above. By the same logic as before, we only need to consider the case where $z$ has order five, and thus so does $z^2$,$z^3$, and $z^4$, so we can write $G$ as:
$$ G=\{ e,x,y,y^2,y^3,y^4,z,z^2,z^3,z^4 \}$$
Finally we consider the element $xy$. By closure, it is either a power of $y$ or a power of $z$ or it is $e$, but if $xy=y^n$ then $x=y^{n-1}$ implies $x$ has order $5$ and likewise if $xy=z^n$ then $x=z^ny^{-1}$ implies $x$ has order $5$, and finally if $xy=e$, then $y=x^{-1}=x$ implies the order of $y$ is 2.
Since none of those statements is possible, it must hold that $G$ is cyclic.
