What is the maximum order of the elements in the group of order ${p^4}$ with 3-generator? Let say we have a group of order ${p^4}$ with 3-generator $\langle x \rangle$, $\langle y \rangle$ and $\langle z \rangle$ where $|\langle x \rangle|={p^2}$, $|\langle y \rangle|=p$ and $|\langle z \rangle|=p$.
Since the order of elements divides the order of groups, thus, the order of elements is either 1, $p$, ${p^2}$ ${p^3}$ or ${p^4}$.
Note that the order of the elements from $\langle x \rangle$ is 1, $p$ or ${p^2}$. Meanwhile, the order of the elements from $\langle y \rangle$ and $\langle z \rangle$ is either 1 or $p$.
There is no elements of order ${p^4}$ because this group is not a cyclic group since cyclic group only have one generator. So, the maximum order of the elements in this group cannot be ${p^4}$.
How about the elements of order ${p^3}$? Is it there is any element of order ${p^3}$? If no, how to show it?
Thank you.
 A: Let $G$ be the group in question. Assume first that $3$ is the minimal number of generators of G. Then $G/\Phi(G)$ is elementary abelian of order $p^3$ and hence has exponent $p$, and $\Phi(G)$ also has exponent $p$ and hence $G$.
$G$ is clearly not cyclic, so the minimal number of generators must be $2$.
If $p=2$, then we could have $G= D_{16}$ dihedral of order $16$, which has exponent $8=p^3$. So when $p=2$  it is possible to have an element of order $p^3=8$. ($G$ could also be the semidihedral group in that case.)
But when $p$ is odd we cannot have element of order $p^3$. For suppose $x$ is such an element. Then $\langle x \rangle$ is a normal subgroup of $G$, and $G = \langle x,y \rangle$ for some $y \in G$.
Now if $G$ was abelian it would have exponent $p^2$ so it must be nonabelian, and hence conjugation by $y$ induces an automorphism of $\langle x \rangle$ of order $p$.
Let $A = {\rm Aut}(\langle x \rangle)$. Then $|A| = p^2(p-1)$, and $A$ has a cyclic Sylow $p$-subgroup generated by $x \mapsto x^{1+p}$. So we can assume that the element of $A$ induced by conjugation by $y$ is $x \mapsto x^{1+p^2}$; in other words, $y^{-1}xy=x^{1+p^2}$.
So $[G,G] = \langle x^{p^2} \rangle$ has order $p$, and we must have $G/[G,G] \cong C_{p^2} \times C_{p}$. Now any generting set of $G/[G,G]$ must include an element of order $p^2$, and all such elements $g[G,G]$ satisfy $g^p[G,G] = x^p[G,G]$ and hence $g^{p^2} = x^{p^2} \ne 1$. So all generating sets of $G$ contain an element of order $p^3$.
