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.


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.

  • $\begingroup$ 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. $\endgroup$ – Derek Holt Sep 25 '15 at 18:59
  • $\begingroup$ 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 $4$. $\endgroup$ – JMoravitz Sep 25 '15 at 19:06
  • $\begingroup$ @JMoravitz Yes I now understand. Any more method to solve the problem then? $\endgroup$ – rama_ran Sep 25 '15 at 19:09
  • $\begingroup$ Kobe's answer below is already very good. It is a straightforward proof by contradiction (or contrapositive depending on your preference) which started by supposing the conclusion is false and then proving that in that case the hypothesis (that the group was non-abelian) must also have been false. $\endgroup$ – JMoravitz Sep 25 '15 at 19:12

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.

  • 1
    $\begingroup$ Alternative to the first part: If there is an element $a$ of order $8$, then $a^2$ has order $4$. $\endgroup$ – Dominik Sep 25 '15 at 19:01
  • $\begingroup$ @Dominik Please explore what you want to say. $\endgroup$ – rama_ran Sep 25 '15 at 19:05
  • $\begingroup$ @rama_ran It's just another way to show that such a group can have no element of order $8$. $\endgroup$ – Dominik Sep 25 '15 at 19:10

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