Let $0\to A\to E\to B\to 0$ be an exact sequence of commutativ groups with $A$ and $B$ finite with orders $a$ and $b$, where $\gcd(a,b)=1$. $B'=\{x\in E|bx=0\}$.
Then is $E=A\oplus B'$ and $B'$ is the only subgroup of $E$ isomorphic to $B$.
Proof:
Since $\gcd(a,b)=1$ there are $r,s\in\mathbb{Z}$ such that $ar+bs=1$. If $x\in A\cap B'$, then $ax=bx=0$, hence $(ar+bs)x=x=0$; and $A\cap B'=0$. Moreover, all $x\in E$ can be written as $x=arx+bsx$; since $bB'=0$, we have $bE\subset A$, hence $bsx\in A$; on the other hand, from $abE=0$ follows that $arx\in B'$. Hence we see that $E=A\oplus B'$ and the projection $E\to B$ defines an isomorphism of $B'$ onto $B$. Conversly, if $B''$ is a subgroup of $E$ isomorphic to $B$, we have $bB''=0$ hence $B''\subset B'$ and $B''=B'$ because these groups have the same order.
I have some questions to steps of this proof. Some are clear, but others I simply do not get.
1.) Why does from $x\in A\cap B'$ follows, that $ax=bx=0$. That $bx=0$ is clear. But why is $ax=0$. I tried using, that the given sequence is exact and I believe that all the steps which I do not get use this fact. But I simply do not see how and where my mistake is.
2.) Why does from $bB'=0$ follow, that $bE\subset A$ and the elements in $A$ are from the form $bsx$?
3.) Why is $abE=0$ and similar to 2.) why is $arx\in B'$.
4.) What is meant with the projection $E\to B$?
I would appreciate a hint on which part of the properties I do not understand, so I can think about the exact arguments myself. Thanks in advance.
by the question you give,the group means additiveq group. 1.since the order of A is n,so nx=0;
2.denote $p:E\rightarrow B$,$p(by)=bp(y)=0,\forall y\in E$,since the order of B is $b$,so $by\in A$. $bE'=0$ I think is superflous.
3.$abE=0$ since $bE\subset A$.
4.remark you can understand $B\cong E/A$,so the projection means $E\rightarrow E/A$