Akababa

Like if you just think of N as the union of orbits of the subgroup generated by those 3 elements

So actually, I think that would be enough to show a isn't in N because you can't find a shorter length sequence anywhere in the orbit

It might help to note that the normality assumption only lets you take $a$ from the front and put $a^{-1}$ in the back

I hope my answer is almost there and maybe you can finish it?

I'm not 100% sure as I didn't actually take group theory yet. I'll have to think about it

Well, i edited it to add in some more details. It's too tedious to put everything in but I hope this is a helpful hint.

Yeah, good point. I think you can still show the structure of $F/N$ with my answer which would trivially imply that $a\not\in N$.

Note that multiplying the generators of $N$ always makes the length longer, because you won't ever have an element ending with $a^{\pm 1,\pm 7},b^{\pm 1, \pm 3}$, so unless you start with $a$ you won't get $a$.

It's really the same thing as $F/N$ is just the cosets of $N$. $N$ is all the stuff you need to multiply $a^{x_1}b^{x_1}...a^{x_n}b^{x_n}$ by to get down to $a^{\{0,1,2,3,4,5,6\}}b^{\{0,1,2,3\}}$.

I don't think that is necessary for the solution. But if $a\in N$ then $F/N$ would collapse down to only powers of $b$.

Hi