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Let p be a prime number and consider $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p^2\mathbb{Z}$. How many subgroups of order p does it have? Given any two subgroups $B_1, B_2 $ of $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p^2\mathbb{Z}$ of order p, is there an automorphism $f$ of $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p^2\mathbb{Z}$ such that $f(B_1) =B_2$?

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Is $p$ prime? Might affect the solution if that is not an assumption. – Thomas Andrews Oct 30 '12 at 16:49
Also, do you really mean one automorphism? Or do you mean if $B_1$ and $B_2$ are subgroups of order $p$ does there exist an automorphism such that $f(B_1)=B_2$? Because your question, as stated, is obviously false if there is more than one subgroup of order $p$, since an automorphism is $1-1$. – Thomas Andrews Oct 30 '12 at 16:52
@ Andrew: You are indeed correct, p is a prime. I will clarify the question immediately. – user44069 Oct 30 '12 at 16:58
@Andrew: I hope it is clear now. – user44069 Oct 30 '12 at 17:00
up vote 1 down vote accepted

Hint: The subgroups of order $p$ are generated by $(1,pk)$ for $k=0,..,p-1$ and $(0,p)$. That gives $p+1$ distinct subgroups.

Now, find two such subgroups, $B_1$ and $B_2$ such that $G/B_1\not\cong G/B_2$ (where $G$ is your product group.) That would answer your second question in the negative.

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Great hint! Thanks a lot! – user44069 Oct 30 '12 at 17:08

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