This is almost a duplicate of the following questions (but, read further):

Group of order $63$ has an element of order $3$, without using Cauchy's or Sylow's theorems

Show any group of order $275$ has an element of order $5$.

But, I wasn't able to apply the proof to my case of a group of order 10 and an element of order 5.

In my case, if I try a proof by contradiction, I'm left with the fact that every element in $G$ is of order 2, but if I divide them into subgroups, every subgroup is of order 2, and has one unique element.

I also tried other methods to prove the theorem, but couldn't come up with anything meaningful.

I'm a beginner in the field, and am not familiar with theorems such as Cauchy's or Sylow's theorem (I'm aware of Lagrange's theorem). Therefore, please try to provide a simple answer, thanks!


Since $|G|=10$, therefore for $g \in G$ the possible orders are $1,2,5,10$. Suppose

  1. There is an element $a \in G$ such that $|a|=10$, then $|a^2|=5$, thus we have an element of order $5$.
  2. Suppose there is an element of order $5$, then we have nothing to prove.
  3. So what if all the elements have order $2$? Then you can show that the group is abelian (small exercise). Now consider distinct elements $a,b \in G$ such that $|a|=|b|=2$, then $|ab|=2$, hence $H=\{e,a,b,ab\}$ will form a subgroup of $G$. But $|H|=4$ and $4 \not| 10$. It violates Lagrange's theorem. So such a $H$ cannot exist. That means not all the elements can have order $2$.

I hope this resolves it within the scope of things you know at this stage.

  • 1
    $\begingroup$ That's ingenious, thanks! The proof is very simple to understand, yet doesn't look simple to come up with. Do you have a tip as of how to approach to tasks like this? $\endgroup$ – Paul May 21 '15 at 21:01

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