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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.

Arturo Magidin
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1 Answers1

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Note that there is a problem with your mapping $\varphi$. It will be an homomorphism provided you know that $G$ is abelian, otherwise it might not be true.

This problem asides, to answer your question, the image of $\varphi$ contains $1,a,\ldots,a^{p-1}$ and $b$ by definition. Using your assumptions on $a$ and $b$, you might be able to show that these elements are parwise distinct.

GreginGre
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