# Abstract: If $G$ is a group of order 10 with only one element of order 2, then $G$ must be cyclic. [closed]

Abstract Algebra:

If $G$ is a group of order 10 with only one element of order 2, then $G$ must be cyclic.

Thank you.

• Hint: There will be only one Sylow-5 subgroup. – User Mar 3 '16 at 5:10
• I can get cyclic $\to$ "only one element of order 2", but how to show from "only one element of order 2" $\to$ cyclic? – Anonymous Mar 3 '16 at 5:16

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.

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.

• @Member thank you, yours is quite nice and concise. – ASKASK Mar 3 '16 at 5:42
• This is a nice proof, but it might be good to note that the reason $z$ and its powers are distinct from $x, y, y^{2}, y^{3}, y^{4}$ is because $\langle z \rangle \cap \langle y \rangle$ has order dividing $5$ by Lagrange, and thus these must be equal (which is not the case by hypothesis) or all elements are distinct, as desired. Similarly, $\langle z \rangle \cap \langle x \rangle = e$, since it must have order dividing $\gcd(2, 5) =1$. – Alex Wertheim Mar 3 '16 at 6:18
• @AlexWertheim good catch. $z$ is different from the others by definition, however I did not consider the case where a power of $z$ may not be different from the others. – ASKASK Mar 3 '16 at 6:20