Approach $196=7^2*2^2$ $Syl_7{G} \equiv 1(mod\text{ } 7) $ and $Syl_7G |4 $, so the only possibility is 1 which implies there is just 1 $Syl_7$ subgroup. This is group is normal in G since every sylow p subgroup are conjugate and there is just one, so we can extablish the following subnormal series
$$1_G \lhd P \lhd G$$
Now we have to show that the quotients are abelian. That's where I got stuck
For primes numbers $p$, any group $G$ of $p^2$ order is abelian. This can be proved by using the conjugacy class equation to rule out $|Z(G)|=1$ and then arguing that each element's centralizer group must be the whole of $G$.
The quotient $P/1\cong P$ has order $p^2$ for $p=7$, and hence is abelian, since every $p$-group has non-trivial center, and groups of order $p$ are cyclic (the center $Z(P)$ must have order $p$, or $p^2$. If $|Z(P)|=p^2=|P|$, we have $Z(P)=P$, and we $P$ is abelian. Otherwise $P/Z(P)$ has order $p$, and hence is cyclic so that $P$ is abelian). Similarly, the quotient $G/P$ has order $2^2$, and hence is abelian (because anyway every group of order $4$ is abelian).
