How does SETH implies ETH?
The more usual (and slightly stronger) formulation of the exponential time hypothesis is that there is $\epsilon &gt; 0$ such that solving 3SAT requires time $\Omega(2^{\epsilon n})$ (see, for example, Abboud, Backurs and Williams), where $n$ is the number of variables.

The strong exponential time hypothesis states that for each $\epsilon &gt; 0$ there exists $k\ge3$ such that solving $k$SAT requires time $\Omega(2^{(1-\epsilon)n})$ (again, see the paper mentioned above).

Showing that SETH implies ETH isn't trivial. The idea is to show that $k$SAT reduces to sparse $k$SAT, which is $k$SAT restricted to instances with $O(n)$ clauses. Such instances can be coded as 3SAT instances with $O(n)$ variables. Therefore if we can solve 3SAT in time $O(2^{\gamma n})$ for every $\gamma&gt;0$, then we can also solve $k$SAT in time $O(2^{\gamma n})$ for every $\gamma&gt;0$. For the details, consult Impagliazzo, Paturi and Zane, Which problems Have strongly exponential Complexity?.