Question

Prove that every non abelian group G has order at least 6; hence, every group of order 2, 3, 4, or 5 is abelian.

Short answer

Every non abelian group has order at least 6.

Step-by-step solution

Elements in H={e,a,b,ab,ba} are all distinct

Let G be a group, $a,b\in G$ and $ab\ne ba$.

It is clear that the elements e, a, b are distinct and $b\ne a^{-1}$ implies $ab\ne e$, $ba\ne e$.

Since, $b\ne e$ then role="math" localid="1654258579993" $ab\ne a$ and $ba\ne a$ . Similarly, $ab\ne b$ and $ba\ne b$.

Thus, all the elements in the subset $H$ are all distinct.

Either a2∉H or  a2=e

Since, $a^2$commutes with a implies $a^2=e$ and $a^2\in H$. Since, e, a, b are all distinct implies $a,a^2,ab,ba$ are all distinct.

If $aba=e$then $ab=a^{-1}=a$ , if $aba=a$then $ab=e$, if $aba=b$ then $ab=b{a}^{-1}=ba$ and if $aba=ba\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}ab$then $a=e$ implies role="math" localid="1654258967789" $aba\notin H$is contrary. Thus, $aba\in H$.

Therefore, every non abelian group has order at least 6.