# Showing that a group with a presentation is free/not free

Show that the group with presentation $\langle a, b, c \mid a^2cb^3\rangle$ is free with basis $\{a, b \}$. Show that the group with presentation $\langle a, b, c \mid a^3b^3 \rangle$ is not free.

I'm not really sure how to get going with this - clearly I need to get rid of the relations but I didn't think Tietze transformations will be able to do that - any help appreciated!

• Did you try using Tietze transformations to remove $c$ from the first presentation? – Ayman Hourieh May 30 '14 at 11:36
• For the first part, note that $c=a^{-2}b^{-3}$. Then, just show that the map $f\colon\langle a,b,c\mid a^2cb^3\rangle\to\langle a,b\rangle$ given by $f(a)=a, f(b)=b, f(c)=a^{-2}b^{-3}$ is an isomorphism. – Dan Rust May 30 '14 at 11:37

Solving the relation for $c$, we conclude that there is a homomorphism $\langle\, a,b,c\mid a^2cb^3\,\rangle\to \langle a,b\rangle$ given by $a\mapsto a$, $b\mapsto b$, $c\mapsto a^{-2}b^{-3}$, which is an isomorphism.

There exists a homomorphism $\langle \,a,b,c\mid a^3b^3\rangle \to \mathbb Z/3\mathbb Z\times \mathbb Z/3\mathbb Z$ given by $a\mapsto (1,0)$, $b\mapsto(0,1)$, $c\mapsto (0,0)$. We conclude that $a\ne b^{-1}$ in $\langle \,a,b,c\mid a^3b^3\rangle$ Let $f\colon\langle \,a,b,c\mid a^3b^3\,\rangle\to\langle u,v,\ldots\rangle$ be a homomorphism into a free group. Then we have $f(a)^3f(b)^{3}=1$ and conclude $f(a)=f(b)^{-1}$. As $a\ne b^{-1}$, we see that $f$ fails to be injective.

• This uses the not completely trivial fact that for elements $u,v$ in a free group, $u^3=v^3$ (or more generally $u^n=v^n$ for some $n \ne 0$) imples $u=v$. – Derek Holt May 30 '14 at 12:00

Here is a different way of doing the second problem. How many homomorphisms are there from $G = \langle a,b,c \mid a^3b^3 \rangle$ to $C_6$ (cyclic group of order $6$)?

We can map $a$ and $c$ to each of the $6$ elements independently, and then we must map $b$ to an element whose cube is the same as the cube of the image of $a^{-1}$. In $C_6 = \langle x \rangle$, there are three elements each with cubes $1$ and $x^3$, so there are three choices for the image of $b$, making a total of $6 \times 6 \times 3 = 108$ homomorphisms.

For a free group of rank $n$, there are $6^n$ such homomorphisms so, since $108$ is not a power of $6$, $G$ cannot be free.

• That's a really nice observation. – Dan Rust May 30 '14 at 13:27
• Alternatively, observe that the abelianization of a free group is a free abelian group, whereas the abelianization of $G$ is ${\mathbb Z} \oplus {\mathbb Z} \oplus {\mathbb Z}/(3{\mathbb Z})$. – Derek Holt May 30 '14 at 13:43