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
Well defined map
3
$\begingroup$
group-theory
-
0http://math.stackexchange.com/questions/45696/prufer-group-is-a-vector-space-over-mathbbz-p-mathbbz – 2011-07-05
-
1Glad 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. – 2011-07-05
-
0Thank 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? – 2011-07-05
-
0@Jyrki: In order to prove that the map is well defined, is it sufficient to prove it for the first $(n+pZ)$? Or is it necessary to make the same proof for $(m/p^i+Z)$ as I tried above? Thanks – 2011-07-05
-
0Yes, you should check that the mapping is independent of the presentation of the element of $H$. Before I can say more, I need to ask you a few things: 1) By Prüfer group do you mean the subgroup of $P\le\mathbf{Q}/\mathbf{Z}$ consisting of elements of prime power orders for a fixed prime $p$? 2) Does the notation $H=G^*[p]$ mean that $H$ is the $p$-torsion part of the group $G^*$? If the answer is 'yes', then doesn't it follow that $i=1$ always? 3) If $G^*$ is a product of, say, $\ell$ copies of the Prüfer group, then aren't the elements of $H$ vectors with $\ell$ components, each in $P$? – 2011-07-05
-
0Ok, but why is $(n+pZ)l$ equal to zero? $G^*$ is a direct sum of a certain number Prufer Groups, that is not $l$, since $l$ is just the integer of our proof. I think there is no connection because by $l$ I mean an arbitrary integer. By Prufer group I mean the $p$- primary component ( that means that orders are some power of the fixed prime $p$) of $Q/Z$ which consists of all cosets $(m/p^i+Z)$, and particularly elements in $H$ are elements of type $(m/p^i+Z)$ with order exactly $p$. – 2011-07-06
-
0What I want to prove is a part of a proof that I studied, so I don't Know If I wrote all importants step.But I am sure that I have to prove that the map is well-defined with this Hypothesis. – 2011-07-06
-
1@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 $$\frac{n'm'}p+\mathbf{Z}=\frac{(n+pk)(m+p\ell)}p+\mathbf{Z}=\frac{nm}p+(mk+n\ell +pk\ell)+\mathbf{Z}=\frac{nm}p+\mathbf{Z},$$ 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. – 2011-07-06
-
1@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. – 2011-07-06
-
0Ah, now I understand what you mean. Thank you much for your help! – 2011-07-07