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My book defines a group $G$ to be the internal direct product of its subgroups $H$ and $K$ if

  1. $H$,$K$ are normal in $G$
  2. $H\cap K=${$e$}
  3. $G=HK$

From these conditions we can prove that $G≈H×K$. This seems much more like a definition of internal direct product to me (I could be mistaken). But while proving this I didn't fully use the condition 1. I merely used a consequence of it, which is that every element of $H$ commutes with elements of $K$. My question is that why did we state a stronger condition? Couldn't we have a situation where $G≈H×K$ but one of $H,K$ is not normal in $G$?

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1 Answer 1

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No, we can't have an isomorphism $G\cong H\times K$ without $H$ and $K$ being normal, for the simple reason that $H$ and $K$, seen as subgroups of $H\times K$ through the canonical inclusions $h\in H\mapsto (h,1)$ and $k\in K\mapsto (1,k)$, are normal in $H\times K$, since they are the kernels of the canonical projections $H\times K\to K$ and $H\times K\to H$, respectively.

In fact, the assumption that $G=HK$ together with the assumption that $H$ and $K$ commute imply that $H$ and $K$ are both normal.

Indeed, if $g\in G$, then you can write $g=hk$ for some $h\in H$ and $k\in K$. Then you have for all $h'\in H$ $$gh'g^{-1}=hkh'k^{-1}h^{-1}=hh'h^{-1}\in H,$$ and thus $gHg^{-1}\subset H$. Since $kh=hk=g$, you can similarly prove that $gKg^{-1}\subset K$, and since these hold for all $g\in G$ both $H$ and $K$ are normal.

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  • $\begingroup$ Sorry for jumping in so late, but by "with the assumption that $H$ and $K$ commute" do you mean $\forall h\in H,k\in K, hk=kh$? $\endgroup$
    – Devo
    Sep 29 at 14:32
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    $\begingroup$ @Devo Yes, that's what I meant. $\endgroup$
    – Arnaud D.
    Sep 29 at 20:15

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