How to show that $⟨a,b | aba=bab⟩$ is not the trivial group?

Question

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

Answer 1

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 $\mathbb Z$.

Answer 2

The deficiency of a finite presentation $<X|R>$ is defined to be $|X|-|R|$. If $\operatorname{def}(G)>0$ then group $G$ is of order infinite.