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Show that $\phi : H\times N\to G$ defined by $\phi (h, n) = hn$ is a injective homomorphism. $H$ and $N$ are normal subgroups of $G$, and $H\cap N = \{e\}$.

I know that if $hn = e$, then $h=e=n$, and $hn=nh$, for $h\in H$ and $n\in N$. I know the definition of homomorphism but I'm having trouble showing in this case.

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    $\begingroup$ So you know $hn=nh $. Now what is the problem? You can write $h_1h_2n_1n_2$ as $h_1n_1h_2n_2$. $\endgroup$
    – Cop 663
    Jul 10 '19 at 2:00
  • $\begingroup$ Can you elaborate it for me? Still don't get it. $\endgroup$
    – Vityôk
    Jul 10 '19 at 2:12
  • $\begingroup$ See This. $\endgroup$
    – Bach
    Jul 10 '19 at 2:17
  • $\begingroup$ You can use MathJax to format your posts. $\endgroup$ Jul 10 '19 at 3:21
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To show $\phi $ is a homomorphism, we need to prove that $\phi ((h_1,n_1)\cdot (h_2,n_2))=\phi ((h_1,n_1))\phi ((h_2,n_2)). $

Note that $$\phi ((h_1,n_1)\cdot (h_2,n_2))=\phi((h_1h_2,n_1n_2))=h_1h_2n_1n_2=(h_1n_1)(h_2n_2)\\=\phi ((h_1,n_1))\phi ((h_2,n_2)).$$

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  • $\begingroup$ To show that ϕ is injective we need to show that ker ϕ = {e}. So Ker ϕ = {h, n ∈ H×N | ϕ(h, n) = e}. Let ϕ(h, n) = hn = e, as we already know this only happens if h=e=n. Therefore, Ker ϕ = {e}. Is this proof correct? $\endgroup$
    – Vityôk
    Jul 10 '19 at 2:59
  • $\begingroup$ @Vityôk Yes, you are right. $\endgroup$
    – Cop 663
    Jul 10 '19 at 3:08

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