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.


It looks like you figured out that the there are 24 elements left with order not $1,5,7$.  That implies the remaining elements must have order 35 (by Lagrange, I guess).
But if $G$ has an element of order $35$, then $G$ is isomorphic to the cyclic group $C_{35}$. Now, note that cyclic groups are indeed abelian.
