Let $G$ a group of cardinality $p^2$ such that every element of $G$ is of order different than $p^2$.
Let $a,b\in G\setminus\{e\}$ such that $b\notin\langle a\rangle$. Of course the order of $a$ and $b$ is $p$.
We consider the mapping $\varphi:(Z/pZ)^2\to G, (\overline{k},\overline{\ell})\to a^kb^\ell$.
Of course we can show easily that this mapping is well-defined and homomorphism of groups.
For the surjectivity of $\varphi$ i have the following indication: Show that $Im(\varphi)$ contain at least $p+1$ element.
If that is true, we can conclude easily that $Im(\varphi)=G$. but the question here why $Im(\varphi)$ contain at least $p+1$ element. Thank you for your help in advance.
\langleand\ranglefor delimiters, not<and>. – Arturo Magidin Sep 15 '24 at 19:00