Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
+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

2 Answers 2

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 $k$.

share|improve this answer
1  
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.

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.