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Suppose $H,N\leq G$, $N\unlhd G$, $H$ and $N$ have trivial intersection, and $HN=G$.

I want to show that if $G\approx H\times N$ (isomorphic), then $H\unlhd G$. What I do is identify $H$ with $H'=\{(h,1)|\ h\in H\}$, and try to show $H'\unlhd H\times N$. I get $(h',n)(h,1)(h'^{-1},n^{-1})=(h'hh'^{-1},1)\in H'$, so it seems like $H'\unlhd H\times N$. I just feel this is taking a leap in logic. Is there a more formal/more correct way to get this implication?

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+1 for showing how you approach the problem. But I'm not convinced that the claim is true. See my answer for details. May be I misunderstood something? – Jyrki Lahtonen Sep 9 '11 at 13:52

Your claim is false as stated. The problem is that the isomorphism $H\times N\rightarrow G$ does not necessarily map elements of the form $(h,1)$ into $H$.

Let me demonstrate this with a counterexample. Let $G$ be the dihedral group of 12 elements that we think of as the group of symmetries of a regular hexagon. Let $N$ be the subgroup of symmetries of a triangle (formed by every other vertex of the hexagon). It is of index 2, so normal in $G$. Then there are altogether six reflections in $G$, and three of those belong to $N$. Let $H\simeq C_2$ be generated by a reflection not in $N$. Then $HN=G$, and $H\cap N$ is trivial. But also $G=NK$, where $K$ is the subgroup generated by the antipodal map (= rotation by 180 degrees). Furthermore, this is actually a direct product $G=N\times K$. $K=Z(G)$ so it, too, is normal, and the intersection $K\cap N$ is trivial, because the 180 degree rotation does not map the triangle back to itself.

Here $K$ and $H$ are isomorphic, because both are cyclic of order two. Therefore also $G=N\times K\simeq N\times H$. Yet $H$ is not a normal subgroup of $G$.

[Edit] A more general scenario, where your claim fails is the following. Let $N$ and $K$ be groups such that there exists a monomorphism $i:K\rightarrow N$ with the property that $i(K)$ is not contained in the $Z(N)$. Let $G=K\times N$. Let $H$ be the subgroup $$ H=\{(k,i(k))\mid k\in K\}. $$ Identify $N$ with $1\times N$. Then $N\unlhd G$. Also $NH=G$, because an arbitrary element $(k,n)\in G$ can be written as a product of an element of $N$ and an element of $H$: $$ (k,n)=(1,n i(k)^{-1})(k,i(k)). $$ Also clearly $H\cap N=\{1_G\}.$ Because the mapping $f:K\rightarrow H, k\mapsto (k,i(k))$ is an isomorphism of groups, we then have $G\simeq N\times H$.

But the subgroup $H$ is not normal. To see this let us pick $k\in K$ and $n\in N$ such that $ni(k)n^{-1}\neq i(k)$. This is possible, because we assumed that $i(K)\not\subset Z(N)$. Choose $h=(k,i(k))\in H$. Conjucating this with $(1,n)$ gives $$ (1,n)(k,i(k))(1,n)^{-1}=(k,ni(k)n^{-1})\neq h. $$ This is not an element of $H$ as $h$ the only element of $H$ with a first component equal to $k$.

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Well-spotted. Sometimes we erase the distinction between isomorphism and equality at our own peril... – Zhen Lin Sep 10 '11 at 10:36

First set up a homomorphism $\theta : H\times N \to N$ by $\theta: (h,n)\mapsto n$.

This has kernel $K_2 = Ker\theta = \{i = (h,n)\in H\times N:\theta(i) = 1\} = \{(h,1):h\in H\}\cong H$.

We then set up an isomorphism $\psi:G\to H\times N$ and look at the composition $\theta\circ\psi$ and try to find $K_1 = Ker\theta\circ\psi = \{g\in G:\theta(\psi(g)) = 1\}$ but since $\psi$ is injective and surjective we get $K_1 \cong K_2$ and is therefore isomorphic to $H$ and we know that the kernel of a homomorphism is a normal subgroup.

I think that comes to the desired conclusion.

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It proves that $G$ has a normal subgroup isomorphic to $H$. This is different from the claim $H\unlhd G$. Let's wait for the OP to clarify, whether this is what she wanted to ask. – Jyrki Lahtonen Sep 10 '11 at 9:17

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