Let $G$ be a finite group such that every Sylow subgroup of $G$ is normal and abelian.Show that $G$ is abelian.
Let $x,y\in G$ .
Case 1:If $x,y$ are in the same Sylow subgroup and as it is given to be abelian so $xy=yx$.
Case 2:If $x,y$ are not in the same Sylow subgroup, so suppose that $x$ is in a Sylow $p$ subgroup of $G$ say $P$ and $y$ is in a Sylow $q$ subgroup of $G$,say $Q$ where $p,q$ are distinct primes.
Now $P,Q$ are normal and hence $xyx^{-1}y^{-1}\in P\cap Q=\{1\}\implies xy=yx$.
Thus $G$ is abelian.
Is the proof correct?Please suggest required edits.
The proof requires a slight fix, because you appear to assume that if you take an element of $G$, it must be in one of the Sylow subgroups. The cyclic group of order $6$ show that this is not the case, unless $G$ is a $p$-group itself.
There are several ways of doing this correctly.
You may first use the fact that for a fixed prime $p$, the Sylow $p$-subgroups form a single conjugacy class. So in your case for each $p$ there is a single Sylow $p$-subgroup $S_{p}$.
Then use the elementary fact that every element of $G$ can be written as a product of elements of prime-power order.
Now you argue like this. Let $x, y \in G$. You know that $x = x_{1} x_{2} \dots x_{n}$ and $y = y_{1} y_{2} \dots y_{n}$, with $x_{i}, y_{i} \in S_{p_{i}}$. You have already proved that the $x_{i}, y_{i}$ commute, so you are done.
If you know direct products, you are really proving that $G$ is the product of the $S_{p}$.