Before anything else : I know there is a very clean proof of this fact and that it is a special case of Cauchy's theorem. Still, I recently read the first time I did this exercise and found out the following proof :
"Let $G$ be a group of even order $2n$. We only need to show that $G$ contains an element of even order.
Let's suppose that $G$ does not contain such element. Let $(H_i)_{i \in [|1;k|]}$ be a family of subgroups of $G$ such that $\forall i \in [|1;k|]$, $\exists x \in G : H_i=<x>$, $\forall i \neq j$, $H_i \cap H_j = \{1_G\}$ and $\bigcup_{i=1}^k H_i =G$. By such hypothesis we get that $o(G)=\sum_{i=1}^k o(H_i)-(k-1)$, i.e. $2n-k=\sum_{i=1}^k o(H_i)+1$.
- If $k \in 2\mathbb{N}$ : $2n+k$ is even. Since $G$ doesn't have any element of even order, $\forall i \in [|1;k|]$, $o(H_i) \in \mathbb{N} \setminus 2\mathbb{N}$. But an even sum of odd numbers is even, so $\sum_{i=1}^k o(H_i)+1$ is odd, which is absurd.
- Else : $2n+k$ is odd. Like before, an odd sum of odd numbers is odd, so $\sum_{i=1}^k o(H_i)+1$ is even, which is still wrong.
Hence, $G$ contains an element of even order, and it follows that $G$ also contains an element of order 2."
My past self did not seem to found anything problematic with this proof, but I do : why does such a family of subgroups exist ? I can't find any intuitive counter-exemple of this, while I don't understand why it could be true.
My first attempt was to use a subgroup of order 6 and a subgroup of order 9 to show that there could be a element of order 3 within them both, but I found it hard to get such a group easily. I then tried to use generators of two different subgroups in order to prove they were equal in the end, but couldn't do a thing with them.
Thanks for your help !