Does SARP imply WARP? and GARP imply SARP?
Yes, SARP implies WARP. Since all choices are singletons under SARP, we abuse notation and write $C$ as a function that takes alternatives as values.
We prove the contrapositive, every violation of WARP in a setting where choices are unique give rise to a violation of SARP. So assume there are sets $B$, $B'$ containing $x$ and $y$ such that $c(B)=x$ and $C(B')=y$. Letting $B_1=B$ and $B_2=B'$, you get a direct violation of SARP (I take it you require $x_i\neq x_{i+1}$ in the definition of SARP).

In the context of budget sets, SARP implies GARP. Let $B_i=\{x\mid p_ix\leq p_ix_i\}$. Then SARP says that if $p_i x_{i+1}\leq p_i x_i$ but $x_i$ is the unique choice from $B_i$ for $i\in\{1,\ldots,N_1\}$, then $p_N x_i&gt;p_N x_N$. So the case $p_N x_i\leq p_N x_N$ can never occur; GARP cannot be violated.

Now GARP as usually formulated is a condition on data sets consisting of price-choice combinations. It is not required that the same price system leads to the same choice, so GARP does not imply SARP.