# Prove that a non-abelian group of order $8$ must have an element of order $4$.

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 $4$"?

• If there is no element of order $4$ then all elements have order $1$ or $2$. Prove that a group with that property must be abelian. – Derek Holt Sep 25 '15 at 18:59
• As for the converse of Lagrange's Theorem, it is false. If a group has order $8$ it is not guaranteed that it must have an element of order $4$. – JMoravitz Sep 25 '15 at 19:06
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.
• Alternative to the first part: If there is an element $a$ of order $8$, then $a^2$ has order $4$. – Dominik Sep 25 '15 at 19:01
• @rama_ran It's just another way to show that such a group can have no element of order $8$. – Dominik Sep 25 '15 at 19:10