Proving that group $G$ of order $|G|=35$ is Abelian This is my outline of proof:

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*By Sylow's theorems, $G$ has two unique subgroups $H$ and $K$ respectively of order $5$ and order $7$ and both are Abelian; as groups of prime order are Abelian

*Next I use the counting argument to say: Taking any $x\in{H}$ and $y\in{K}$ such that $x,y\notin{e}$, I will have more number of ordered pairs than the remaining elements in $G$ (i.e $35-5-7+1=24$)  ($e$ is common to both so $+1$)

*Thus there are combinations which are same; implying $x$ and $y$ commute; but how do I proceed to prove that even the other elements commute. Kindly help in showing the way further.


If $|G|=35$, then all of it elements must have order $1,5,7$ or $35$ by Lagrange.
The Sylow theorems shows that we have one subgroup $H$ of order $5$ and one subgroup $K$ of order $7$. Hence, we have $6$ elements of order $7$ and $4$ elements of order $5$.
So, by counting, $35 - 4 - 6 - 1 = 24$ elements of $G$ must have order $35$, then $G$ is cyclic, hence it's abelian.
