by

Counting by order of elements theorem since $\displaystyle n$ is a divisor of $\displaystyle n$ $\displaystyle \implies\$ number of elements in G of order n is a multiple

of $\displaystyle \varphi($$\displaystyle n)$ $\displaystyle \implies\$ $\displaystyle n=k*\varphi(n)$ for some $\displaystyle k=$positive integer.

gcd($\displaystyle n,\varphi(n)$)=gcd($\displaystyle k*\varphi(n),\varphi(n)$)=1 for some $\displaystyle k=$positive integer.

$\displaystyle \implies\$ $\displaystyle \varphi($$\displaystyle n)=1$ $\displaystyle \implies\$ $\displaystyle n=1$ or $\displaystyle n=2$

if ord(G)=1 then G={e} $\displaystyle \implies\$ G is abelian

if ord(G)=2 then G is abelian since if a,b in G then a*b=b*a because (a=b) or (either a or b equal e).

is this correct?