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I encountered this problem in Gallian Abstract Algebra (pg. 239, 8th edition)

Suppose that $G=H\times K$ and that $N$ is a normal subgroup of $H$. Prove that $N$ is normal in $G$.

Where $\times$ denotes the internal direct product. I'm having some trouble proving this. If $G=H\times K$, we have that $H,K$ are both normal in $G$. For a while I was working under the assumption that $N$ is normal in $G\iff N$ is normal in $H$ and $N$ is normal in $K$, but I'm not sure that this is the case any more. Or perhaps this is the case only if $N=\{e\}$? Am I overlooking something simple? Any help is appreciated! Thank you.

EDIT: I was overlooking the basic result of $G=H\times K$ that every element of $H$ and $K$ commute. Thanks for the help!

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  • $\begingroup$ You could think of it as $N \times 1$ $\endgroup$ – shobon Apr 28 '13 at 20:54
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    $\begingroup$ @julien I think that $N\subseteq G$ . This is an internal direct product $\endgroup$ – Amr Apr 28 '13 at 21:07
  • $\begingroup$ @Amr Oh, I did not see that...Thank you. $\endgroup$ – Julien Apr 28 '13 at 21:07
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Use $N\cong N\times 1$.

$(h,k)^{-1}\cdot (n,1)\cdot (h,k)=(h^{-1},k^{-1})\cdot (n,1)\cdot (h,k)=(h^{-1}nh,k^{-1}1k)=(n',1)\in N$, where $n'=h^{-1}nh\in N$ by normality of $N$ in $H$

$\textbf{EDIT}$: Why do $hk=kh$ for every $h\in H, k\in K$?

$hk=kh$ if and only if $k^{-1}hkh^{-1}=e$. Set $u:=k^{-1}hkh^{-1}$. You have $$u=(k^{-1}hk)h^{-1}=h'h^{-1}\in H$$ where $h'=k^{-1}hk$ is in $H$ by normality of $H$, but also $$u=k^{-1}(hkh^{-1})=k^{-1}k'\in K$$ where $k'=hkh^{-1}$ is in $K$ by normality of $K$ in $G$, Hence $u\in H\cap K=e$

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  • $\begingroup$ And can you show this using that $\times$ is an internal rather than an external direct product? $\endgroup$ – luke Apr 28 '13 at 21:19
  • $\begingroup$ @zach internal and external are isomorphic, provided the intersection of normal subgroups is trivial $\endgroup$ – Federica Maggioni Apr 28 '13 at 21:20
  • $\begingroup$ Right, but is there a way to show it without using that isomorphism? That is, only using the properties of internal direct products. $\endgroup$ – luke Apr 28 '13 at 21:23
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    $\begingroup$ @zach, since $K\cap H=\{e\}$ and both $K$ and $H$ are normal in $G$ you have $kh=hk$ for all $h\in H$ and $k\in K$, then $(hk)^{-1}=k^{-1}h^{-1}=h^{-1}k^{-1}$ $\endgroup$ – Camilo Arosemena-Serrato Apr 28 '13 at 21:36
  • $\begingroup$ @zach I have added a little edit to explain why $H$ and $K$ commute $\endgroup$ – Federica Maggioni Apr 28 '13 at 21:54
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Normality is not transitive: a normal subgroup of a normal subgroup need not be normal in the large group.

For instance, consider the dihedral group $D_8$ of a square's isometries. Then the subgroup $N$ generated by vertical and horizontal reflections is normal, and the group $H$ generated by the horizontal reflection is normal in $N$, but is not normal in $G$. Of course, this is not a counterexample to what you're supposed to show: it just shows that $N$ is not a direct factor of $D_8$.

Instead, I would advise you to just show the normality directly, by the definition. It is very easy to do in this case (note that to show normality it is enough to show invariance under conjugations by elements of some generating set).

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  • $\begingroup$ could you show these easy steps? I'm having trouble proving this directly. $\endgroup$ – luke Apr 28 '13 at 21:17
  • $\begingroup$ @Zach: conjugate $N$ by any element of $H$ and then by any element of $K$. $\endgroup$ – tomasz Apr 28 '13 at 21:19
  • $\begingroup$ $khnh^{-1}k^{-1}=kn_1k^{-1}$ for some $n_1\in N$, but now what? We don't know yet that $N$ is normal in $K$. $\endgroup$ – luke Apr 28 '13 at 21:26
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    $\begingroup$ @Zach: $kn_1k^{-1}=n_1kk^{-1}=n_1$, since $n_1$ is a member of $H$. $\endgroup$ – tomasz Apr 29 '13 at 3:49
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Hint. Don't try any fancy stuff. Just compute $g^{-1}ng$ for an arbitrary $g\in G,n\in N$ and see if it's in $N$. Remember that you can write $g$ as $(x,y)$ and $n$ as $(n,1)$.

$$(x,y)^{-1}(n,1)(x,y)=(x^{-1},y^{-1})(n,1)(x,y)=(x^{-1}nx,y^{-1}y)=\ldots \text{ now you finish.}$$

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  • $\begingroup$ +1 for don't try any fancy stuff. I consider this answer better than mine. $\endgroup$ – Amr Apr 29 '13 at 17:55
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Hint: Consider the kernel of the homomorphism $\phi:G\rightarrow (H/N)\oplus K$ that sends $h+k$ to $(h+N,k)$.

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I'll try this kind of demonstration:

if we prove that $N \unlhd H \Longrightarrow N \unlhd K$ we have proved the thesis because, for every $g=hk \in G$ (by hypothesis we can write every element of $G$ in this way) we have, $\forall n \in N, \ hk \in G$ $$hk \ n \ k^{-1}h^{-1} = h \ \tilde{n} \ h^{-1} \in N$$ So the proof is complete. Now we demonstrate the hypothesis we use ($N \unlhd H \Longrightarrow N \unlhd K$)

A consequence of $G=HK$ (internal direct product) is that every element of $H$ commute with every element of $K$. In $k \ n \ k^{-1}$ note that $n \in H $ so i can commute the two elements in this way $k \ k^{-1} \ n = 1_{g} \ n = n \in N$ so it is normal.

(I think i can assume (in this last part of dim) that $N \leq H$ instead of normality.

I'll prove here that every elements of $H$ commute with every element of $K$ in two ways,

1) by definition of internal direct product of groups otherwise

2) by a consequence of the equivalent definition of internal direct products. we'll prove 2): by hypothesis I have $H \unlhd G, \ K \unlhd G$ $G=HK$ and most important $H \cap K =\lbrace 1_{G} \rbrace$ so $$hk=kh \Longleftrightarrow k^{-1}hkh^{-1} =1_{G}$$ using normality of $H$ and $K$ it is easy to prove that $k^{-1}hkh^{-1} \in H \cap K $ so therefore it must be $1_{G}$ and this conclude the proof.

Hope I haven't done mistakes ^^

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  • $\begingroup$ Why can we say that every element of $H$ commutes with every element of $K$? I haven't learned this as a consequence of the internal direct product. Is this a direct result? $\endgroup$ – luke Apr 28 '13 at 21:35
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    $\begingroup$ There is a list here of property of the direct internal product, if it is this what you mean groupprops.subwiki.org/wiki/Internal_direct_product $\endgroup$ – Riccardo Apr 28 '13 at 21:36
  • $\begingroup$ thank you! I overlooked this basic result. $\endgroup$ – luke Apr 28 '13 at 21:38
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    $\begingroup$ I'm looking for a simple proof of this so i can add it to the answer^^ No problem, your welcome $\endgroup$ – Riccardo Apr 28 '13 at 21:43
  • $\begingroup$ I would appreciate that also. This result makes sense but it is not immediately obvious. $\endgroup$ – luke Apr 28 '13 at 21:45

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