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Show that group $A = \mathbb Z_p \oplus \mathbb Z_p \oplus \mathbb Z_p$ is not generated by two elements. ($p$ is prime.)

any help or hint will be appreciated.

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    $\begingroup$ The group $A$ is not isomorphic to $G=Z_{p^2} \times Z_p$, since $G$ has elements of order $p^2$, but $A$ doesn't. $\endgroup$ – Watson Dec 27 '15 at 21:03
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    $\begingroup$ Hint: let $a,b$ be any two elements. They have order at most $p$. How large can the subgroup generated by them be? $\endgroup$ – Wojowu Dec 27 '15 at 21:04
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    $\begingroup$ BTW, can you prove $\mathbb{R}^3$ has no spanning set of size two? (As a real vector space, not as an additive group.) $\endgroup$ – arctic tern Dec 27 '15 at 21:04
  • $\begingroup$ deleted the mistake . thanks $\endgroup$ – user101010 Dec 27 '15 at 21:06
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    $\begingroup$ Note: the fact that two elements of order $p$ cannot generate a group of order greater than $p^2$ depends on the fact that the group is abelian, which is obvious from the structure, but is not explicit in the proofs (which also work within abelian contexts). To see this note that the symmetric group on three elements of order $6$ is generated by any pair of elements of order $2$ and the alternating group on four elements of order $12$ can be generated by a pair of elements of order $3$. This comment is to highlight that abelian methods may fail in non-abelian contexts. $\endgroup$ – Mark Bennet Dec 27 '15 at 21:43
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You're wrong for the isomorphisms.

Hint:

$\mathbb Z_p\oplus\mathbb Z_p\oplus\mathbb Z_p$ is a $\mathbb Z_p$-vector space. What is its dimension?

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    $\begingroup$ This feels more like a rewording of the problem than a solution... $\endgroup$ – A.P. Dec 27 '15 at 21:09
  • $\begingroup$ 3 dimensions . so can I say that (1,0,0) , (0,1,0) , (0,0,1) will always be in the groups and I can't make them from any two elements because ill always have one of the three elements to be 1 or 0 not both? $\endgroup$ – user101010 Dec 27 '15 at 21:10
  • $\begingroup$ I didn't intend to give a solution, but a hint. The problem comes down to a problem of vectors spaces, not groups. This observation gives new tools to solve the problem. $\endgroup$ – Bernard Dec 27 '15 at 21:12
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    $\begingroup$ @gpgpgp: More simply if $2$ elements generate the group, they generate it as a vector space, hence the dimension would be at most $2$. $\endgroup$ – Bernard Dec 27 '15 at 21:14
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    $\begingroup$ The invariant basis number is non-trivial but it remains a basic fact about finite dimensional vector spaces. One cannot always reinvent the wheel. That said, the combinatorial argument is a fine argument. $\endgroup$ – Bernard Dec 27 '15 at 22:20
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This hint is in the comments, but let's say $x,y$ are two elements in $\mathbb{Z}_p\oplus\mathbb{Z}_p\oplus\mathbb{Z}_p$. The subgroup generated by them is given by all elements of the form $$ax+by$$ where $a,b\in\mathbb{Z}$ (written multiplicatively this is more obvious as it becomes $x^ay^b$). There are $p$ distinct possibilities for $a$ and $p$ distinct possibilities for $b$ (since $x$ and $y$ have order $p$). How many elements does this generate? If it's less than $p^3$ then two elements cannot generate the whole group.

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