Let $G$ be an abelian group of order $6$. Prove that it is impossible that every non-identity element of $G$ has an order of $2$.
(Hint: Construct a subgroup of order $4$ and get a contradiction.)
What i tried:
Following the hint, I know that if i can construct a subgroup of order $4$ it would be a contradiction straight away because by Lagrange theorem, the order of a subgroup must be a divisor of the order of a group and $4$ is not a divisor of $6$.
I let $H$ represent the subgroup and that $H$ has an order of $4$ and thus have $4$ elements. Also every non identity element of $H$ must have an order of $2$.
The subgroup that i have is $H=<1,r,m,rm>$
where $r$ stands for rotation and $m$ stands for reflection. Am i correct? Could anyone explain. Thanks
Hint: Let $x,y$ be different elements of the group (not equal to the identity). Can you show that $\{e,x,y,xy\}$ forms a subgroup?