TITLE: on finitely generated non-abelian p-group
QUESTION [1 upvotes]: Let $G$ be a finitely generated non-abelian p-group for example $G=\langle x,y,z\rangle$
is following argument right?
that mean for every $g\in G$ there is i,j,k 
$g=x^iy^jz^k$ since G is non abelian $g\not =y^jx^iz^k$ then we must have
$G\not =\langle y,x,z\rangle $

REPLY [2 votes]: No, this is not correct. First of all, I'm going to assume that $x, y, z$ are elements of some other known group so that $\langle x, y, z \rangle$ makes sense in the first case. Then $G = \langle x, y, z \rangle$ consists of all elements of the form $x^{i_1} y^{i_2} z^{i_3} x^{i_4} y^{i_5} z^{i_6} x^{i_7} \ldots z^{i_k}$ for any integers $i_1, \ldots, i_k$ and any $k > 0$. Since the exponents are allowed to be zero, it's easy to see that this is equivalent to saying that $G$ consists of all strings consisting of $x, y, z$ and their inverses. So, for instance, $y^jx^iz^k \in G$.
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