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Let $G$ be an arbitrary group with identity element $e$ and let $K$ and $N$ be normal subgroups of G with $K∩N={e}$.

I know that $nk=nk$ And the representation of an element in $K×N$ in form $kn$ is unique for all $k∈K$ and $n∈N$.

How can I show the function

$φ:(K×N)→(K⊕N)$ be defined by $φ(kn)=(k,n)$.

is well defined?

I know that I need to show for all $kn,xy∈K×N$ such that $kn=xy$, $φ(kn)=φ(xy)$ But I don't know how to show that

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I think your third line should begin with "I know that $\,nk=kn\,$ ..." , and not what you wrote... –  DonAntonio Apr 20 '13 at 14:01
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2 Answers 2

up vote 2 down vote accepted

If $kn=xy$, then $x^{-1}k = yn^{-1}$. Since the left hand side is in $K$ and the right hand side is in $N$, both are in $K\cap N=\{e\}$. Hence $x^{-1}k=e=yn^{-1}$ and we conclude $x=k$ and $y=n$.

Thus, $\phi(xy) = \phi(kn)$ by the substitution property of equality.

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\begin{alignat*}{4} &\text{If}\quad &&k_1n_1=k_2n_2 \\ &\Rightarrow {} &&k_2^{-1}k_1=n_2n_1^{-1}=e \quad\quad&&(K\cap N =\{e \}) \\ &\Rightarrow {} &&k_1=k_2 \land n_1=n_2 &&\text{}\\ &\Rightarrow {} &&(k_1,n_1)=(k_2,n_2) &&\text{}\\ &\Rightarrow {} &&\varphi(k_1n_1)=\varphi (k_2n_2) &&\text{}\\ \end{alignat*}

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