TITLE: Showing that a group with a presentation is free/not free
QUESTION [5 upvotes]: 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!

REPLY [7 votes]: 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.