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

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

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

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 $$\mathbb Z$$.

• Also $S_3$ with $a=(1\ 2)$ and $b=(2\ 3)$.
– bof
Dec 7 '18 at 14:41
• @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
– R.P
Dec 7 '18 at 15:38
• If a set surjects onto an infinite set, it is obviously infinite.
– lulu
Dec 7 '18 at 19:50

The deficiency of a finite presentation $$$$ is defined to be $$|X|-|R|$$. If $$deg(G)>0$$ then group $$G$$ is of order infinite.