Sylow subgroup of group of order 56

Let $\displaystyle G$ be a group of order 56, $\displaystyle T \in \text{Syl}_2(G), S \in \text{Syl}_7(G)$ be given. Then $\displaystyle |T| = 8, |S| = 7$. Assume $\displaystyle S$ is not normal, so then $\displaystyle T \lhd G$. Show that $\displaystyle T \cong C_2^3$.

Here's what I have so far. My idea is to show that each element of $\displaystyle T$ has order dividing 2. Let $\displaystyle z \in S$ be given such that $\displaystyle \langle z \rangle = S$. Then $\displaystyle G / T = \{T z^i : i=0,\dots,6\}$ and $\displaystyle \forall x \in T, (Tz)^x = Tz$. I've also got that $\displaystyle G = T \rtimes S$, although I fail to seen how any of this is helping.

Any advice?