I've got $H = \langle(2\, 1\, 3\, 4\, 5\, 6)(7\, 8\, 10\, 9)\rangle$ a subgroup of the symmetric group $S_{10}$. How can I know if the left cosets $(7\, 8)H$ and $(8\, 10\, 9)H$ are equal?
I've calculated all the elements of $H$, and then I've multiplied $(7\, 8)$ with all the elements of $H$; then, I've done the same with $(8\, 10\, 9)$. Like this, I've seen that they are not equal.
But my question is, is it enough to say that: We know that two left cosets are equal or disjoint; so as $(7\, 8){\rm id}$ is into $(7\, 8)H$ but is not into $(8\, 10\, 9)H$, can I say that they are not equal?
Yes, you definitely can. Note that $(7\,8)\mathrm{id}=(7\,8)$ itself. And I presume that you can confidently say that $(7\,8)\notin(8\,10\,9)H$ simply because you've explicitly calculated all elements of $(8\,10\,9)H$, right?
Here's another possible way to look at it. Note that $$xH=yH \iff y^{-1}xH=H \iff y^{-1}x\in H.$$ This criterion is usually stated in abstract algebra courses and textbooks. Since you've already calculated all elements of $H$ explicitly, you can now calculate $(7\,8)^{-1}(8\,10\,9)$ and check if it's in $H$ or not.