Groups of order 6 Show, without Lagrange theorem, that every group $G$ of order $6$ has an element of order $3$. 
I know that if $G$ has an element $a$ of order 6, then $a^2$ has order $3$, but I am not shure how to show that for the case where $G$ does not have an element of order $6$. I also know that we can only have elements $1,2,3,$ or $6$ (we proved this independent without Lagrange theorem).
 A: Let $a\in G$. If $a\ne1,$ then $a$ has order $2,3,6$ (as you said). If $a$ has order $6$ then (as you said)  $a^2$ has order $3$. If $a$ has order $2$ then we only have two elements in $G$. So let $b\in G$ (not equal to either $a$ or $1$) with order $2$. Note that if $(ab)^2=abab=1,$ then $a(abab)b=a1b\implies ba=ab.$ So we have only $\{1,a,b,ab\}$. 
Now add $c$ (suppose it has order $2$ and is not already in $G$). We have at least $5$ elements now: $\{1,a,b,ab,c\}$. We want to show that there are at least $2$ more. Now if it has order $2,$ then $ac$=$ca$ (by the same argument for $ab$). Clearly $ac\ne 1,a,ab,c$.  Suppose that $ac=b$. Then (since $a=a^{-1}$) $c=ab$ but $c\ne ab$. Similarly we can show that $bc$ is not equal to any element already in $G$. So $G$ contains at least $1,a,b,c,ab,ac,bc$.
A: If all nonidentity elements of a group have order $2$, then the group is Abelian. You can then show that any finite abelian group where every nonidenity element has order $2$ must have order a power of $2$.
