I have a problem in understanding one line of "Combinatorial Group Theory" by Magnus, Karrass and Solitar (Page 27-28). (Please find it in the image below)
In equation (4), it considers a group $G=\langle a,b,c\mid a^{-1}ba=c,a^{-1}ca=b,b^{-1}ab=c,a^{-1}ca=b, c^{-1}ac=b,c^{-1}bc=a\rangle$.
In equation (5), it first shows that all words of the form $a^{2k},a^{2k}a,a^{2k}b,a^{2k}c,a^{2k}ab,a^{2k}ba$, ($k$ is integer) are the representative for the equivalence class of (4). Then, it claims that no two of the canonical word in (5) are equivalent. For this, it first defines a homomorphism from $G$ to the symmetric group of order $6$ and then it defines a homomorphism from $G$ to the infinte cyclic group generated by $x$, given by $a \mapsto x, b \mapsto x,c \mapsto x$. It says "we see that equivalent words in (5) must have the same $k$. Therefore, distinct words in (5) are inequivalent."
I don't understand the role of homomorphism from $G$ to infinite cyclic group to conclude this claim (in bold and italic above).
This answer is not a comment, only because i need the space. There is no further mathematical argument inserted, the propositions in the picture are slightly expanded.
So we have to show that no two different words (taken from the free group in the letters $a,b,c$) in the list $$ a^{2k},\ a^{2k}a,\ a^{2k}b,\ a^{2k}c,\ a^{2k}ab,\ a^{2k}ba\ ,\qquad k\in\Bbb Z\ , $$ become equivalent when taken modulo the given presentation. For this, rewrite the list of words in the form $$ a^{2k}\cdot u\ ,\qquad k\in\Bbb Z\ ,\ u\in\mathcal U:=\{1,a,b,c,ab,ba\}\ . $$ Two "different" words are of the shape $$w=a^{2k}u\ ,\ W=a^{2K}U\ ,$$ where the pairs $(k,u)$, $(K,U)$ are different (as pairs), $k,K\in \Bbb Z$, $u,U\in\mathcal U$.
So let us start with $w,W$ as above, we assume they are equivalent as words, i.e. equal in $G$, $w\sim W$, and show that they coincide as words, i.e. $k=K$ and $u=U$.
Equal elements in $G$ are mapped by a morphism also into equal elements. So the relation $w= W$ in $G$ is "mapped" into $k= K$. We may now, and do assume now, $k=K=0$ ( after we possibly simplify with $a^{2k}$ in $w=W$).
Finally, we have to show that the $\mathcal U$-component in $w,W$ is the same one, i.e. that from $u=U$ in $G$ we also have $u=U$ in the free group $\langle a,b,c\rangle$ generated by $a,b,c$. We use now the "test morphism" from $G$ to $S_3$. It is a morphism, because the relations in the presentation are mapped to the trivial permutation in $S_3$. It maps the six elements of $\mathcal U\subset \langle a,b,c\rangle$ in six different permutations in $S_3$. This finishes the argument.
(The long full composition $$\mathcal U\overset\subset\to \langle a,b,c\rangle\to G\to S_3$$ is an injective map, so putting it the other way, if we start with $u'\ne U'$ in $\mathcal U$, and have $u'=U'$ in $G$, then we have also equal images of them in $S_3$, contradiction to the injectivity.)