Prove that a non-abelian group of order $8$ must have an element of order $4$.
To solve it, one can use the concept upto Lagrange's th.
Attempt:
We have $o(element)|o(Group)$ then here order of element is $4$ and order of group is $8$ and $4|8$, then can we generally say that "group of order $8$ must have an element of order $4$"?
Please suggest.
Best Answer
If your group $G$ of order $8$ has no elements of order $4$, then either it has an element of order $8$ (so $G$ is cyclic, in particular abelian) or every nonidentity element of $G$ has order $2$; in the latter case, $(xy)^2 = e = x^2y^2$ for all $x,y\in G$, so $G$ is abelian.