Consider the following presentation for $A_4$ $$< p, q\, |\, p^2 = (pq)^3 = q^3 =e>$$ There exist eight elements of order $3$ in this group. Deduce these elements by writing them as products of $p$ and $q$.
Attempt:
The eight elements I found were $qpq,\, pqp,\, pqpq, \,q^2, \,qp,\, pq^2,\, pq,\, q$. Is that correct? If I take $p = (14)(23)$ and $q = (234)$ then I can generate all $3$-cycles in $A_4$.
In particular, if I take $qp = (234)(14)(23) = (412)$ then this means $(qp)^3 = e$ since $qp$ is a $3$-cycle. However, using just the symbolic notation, I have that $(qp)^3 = q^3p^3 = p^3 = (p^2)p = e \cdot p = p \neq e$. Why is this?
Many thanks.
Your error is in assuming $(qp)^3 = q^3p^3$. However, we can only say
$$(qp)^3 = qpqpqp$$
$(qp)^3= q^3p^3\;$ ONLY if $p$ and $q$ commute, and they do not here.