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I want to show that G = $⟨a,b | aba=bab⟩$ is not the trivial group

I tried to find homomorphism $\phi$ from $G$ to $\mathbb Z$ which maps $a$ to $0$ and $b$ to $1$ (or $b$ to $0$ and $a$ to $1$) and if such a homomorphism exists , $\phi(b)$ is non-trivial and thus b is non-trivial. but I didn't found such a homomorphism.

I'm also tried to conclude it directly from the relation and I failed again

Thanks

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    $\begingroup$ If you add the relation $ab=ba$ then you quickly deduce $a=b$, so there is a surjective map to $\mathbb Z$. $\endgroup$
    – lulu
    Dec 7 '18 at 13:13
  • $\begingroup$ You can define your homomorphism to ${\mathbb Z}$ by mapping $a$ and $b$ to $1$. $\endgroup$
    – Derek Holt
    Dec 7 '18 at 15:12
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Note that $\mathbb Z$ satisfies this, with $a=b$.

Phrased differently, adding the relation $ab=ba$ we quickly deduce that $a=b$, so the new relation gives a surjective map to $\mathbb Z$.

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  • $\begingroup$ Also $S_3$ with $a=(1\ 2)$ and $b=(2\ 3)$. $\endgroup$
    – bof
    Dec 7 '18 at 14:41
  • $\begingroup$ @lulu OK, so we can "extend" $G$ with another relation and get a presentaion of $Z$ , can you please explain how this helps in my case ? Tnx $\endgroup$
    – R.P
    Dec 7 '18 at 15:38
  • $\begingroup$ If a set surjects onto an infinite set, it is obviously infinite. $\endgroup$
    – lulu
    Dec 7 '18 at 19:50
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The deficiency of a finite presentation $<X|R>$ is defined to be $|X|-|R|$. If $deg(G)>0$ then group $G$ is of order infinite.

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