# Well defined map

In the past I have asked this question by giving the wrong hypothesis.So now I'll try to give the right information. I will be grateful for any help. I have to prove that the following map from $Z/pZ×H$ to $H$ defined as: $(n+pZ)(m/p^i+Z)↦(nm/p^i+Z)$ is well defined (is independent of the choice of representatives). Where $H=G^*[p]$ and $G^*$ is a direct sum of Prüfer groups, so elements in $H$ are elements of $G^*$ such that their order is $p$. thanks

-
Glad to see that you sorted this one out. All we need here is $H$ to be an abelian group such that $px=0$ for all $x\in H$. If $n+p\mathbf{Z}=n'+p\mathbf{Z}$, then $n'=n+p\ell$ for some integer $\ell$. But then $$n'x=(n+p\ell)x=nx+(p\ell x)=nx+p(\ell x)=nx+0=nx,$$ for all $x\in H$. I am unfamiliar with your notation $G^*[p]$, so I am hesitant to write this as an answer. As long as the statement about the order of elements of $H$ is true, this is well defined. – Jyrki Lahtonen Jul 5 '11 at 6:48
Thank you Jyrki. For the second part : if $(m/p^i+Z)=(m'/p^i+Z)$, then $m'/p^i=(m/p^i+l)$ for some integer $l$. But $(n+pZ)(m'/p^i)=(n+pZ)(m/p^i+l)=(n+pZ)(m/p^i)+(n+pZ)l$. So $(n+pZ)l$ should be equal to zero? – stacy Jul 5 '11 at 8:07
@stacy: I'm guessing that you are supposed to draw a conclusion that because $m/p^i+\mathbf{Z}$ is a $p$-torsion element (assuming that this is what $G^*[p]$ means), then we must have $i=1$. In that case everything works out fine: if $n+pZ=n'+PZ$ and $m/p+\mathbf{Z}=m'/p+\mathbf{Z}$, then $n'=n+pk$ and $m'=m+p\ell$, and $$\frac{n'm'}p+\mathbf{Z}=\frac{(n+pk)(m+p\ell)}p+\mathbf{Z}=\frac{nm}p+(mk+n\el‌​l +pk\ell)+\mathbf{Z}=\frac{nm}p+\mathbf{Z},$$ and the mapping is well defined. But I don't know for sure, whether this is was $G*[p]$ means. Only you can tell that. – Jyrki Lahtonen Jul 6 '11 at 16:23
@stacy: And my other question was about the $G^*$ being a product of copies of Prüfer groups. Then an element of $G^*$ looks like $$(\frac{m_1}{p^{i_1}}+\mathbf{Z},\frac{m_2}{p^{i_2}}+\mathbf{Z},\ldots,\frac{m_‌​\ell}{p^{i_\ell}}+\mathbf{Z}).$$ Not just a single component. Mind you, this won't really complicate your problem as long as we can understand the part about $H$ only having $p$-torsion elements. – Jyrki Lahtonen Jul 6 '11 at 16:32