Free Group where generators aren't just one symbol?

Question

The question is:

Show that $G = \langle a, b, c \mid a^2bacacab \rangle$ is a free group on the free generators $ab$ and $ac$.

This is in section 1.4, page 39 of Combinatorial Group Theory by Magnus, Karass, and Solitar (standard graduate text on the subject), where it is mentioned but not proved or given as an exercise.

I can see by straightforwardly manipulating this relator that indeed $a$, $b$, and $c$ can each be expressed as a product of $ab$, $ac$, and their inverses (for example, c = abacacabac).

However,

1. How do I make sense of "the free group on $ab$ and $ac$." Does this mean that $G$ is isomorphic to $\langle d,e \rangle$ under the rewriting rule $ab \to d$, $ac \to e$? This needs to be proved separately, no?
2. Why should the free group on $ab$ and $ac$ satisfy $G$'s relator?

If it helps, I tried the following on GAP:

F := FreeGroup(3);
a := F.1;
b := F.2;
c := F.3;
G := F/[(a^2)*b*a*c*a*c*a*b];
StructureDescription(G);
    Error, no method found!
IsFreeGroup(G);
    false

So structure description doesn't work here, and it says it's NOT a free group? Maybe that's just a technicality where IsFreeGroup(G) only tells you if G was entered exactly as a free group.

Please let me know if you can help me with any of the above, or just give me an independent proof without reference to what I've written about my progress thus far. Thank you.

Answer 1

If $x$ and $y$ are elements of a group $G$, then when we say $G$ is free on $x$ and $y$, we mean that the unique homomorphism from the free group on indeterminates $X$ and $Y$ that maps $X$ to $x$ and $Y$ to $y$ is an isomorphism.

The relator tells you that $a(ab)(ac)^2(ab) = 1$ so that $a = ((ab)(ac)^2(ab))^{-1}$. So you you can write $a$ as $(xy^2x)^{-1}$, where $x = ab$ and $y = ac$ and so eliminate $a$ from a presentation of $G$ in terms of $a$, $x$ and $y$. Can you take it from there?

Answer 2

I see it now. Thank you for clarifications from others.

$\langle a,b,c|a^2bacacab \rangle \cong \langle a,b,c,d,e|d=ab, e=ac, a^{-1} = de^2d \rangle \cong \langle a,b,c,d,e | b=de^2d^2, c=de^2de, a^{-1} = de^2d \rangle \cong \langle <d,e|>\rangle$

which shows that indeed $ab$ and $ac$ make a free group, as well as the original question that this is isomorphic to G.