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If $N,K$ are subgroups of a group $G$ such that $G=N \times K$ (meaning $G$ is isomorphic to $N \times K$, and the book also says they use this notation when $N,K$ are normal subgroups of $G$ and each member of $G$ can be writen uniquely as the product of some $n_1m_1$) and $M$ is a normal subgroup of $N$, prove that $M$ is a normal subgroup of $G$.

Well, $M \leq N \leq G$ so $M \leq G$. As far as showing normality of $M$ in $G$ I am either thinking about showing if $g \in G$ then $g^{-1}Mg \subset M$ or going the route of cosets and showing $gM = Mg$.

If $g \in G$, then by hypothesis $g$ can be uniquely written as $g=nk$ for some $n \in N$ and $k \in K$. Since each element of $g$ can be written uniquely it follows that $N \cap K = \{ e \}$, trivial subgroup, and $N,K$ are normal so $g=nk=kn$. I haven't gotten very far by the conjugation route nor with the coset route. Any help or see anything wrong with my thinking so far? Appreciate it, cheers!

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Note that the relation of being a normal subgroup is not transitive in general. For example, inside the symmetric group, the group $V = \langle 1, (12)(34), (13)(24), (14)(23) \rangle \triangleleft A_4 \triangleleft S_4$, and yet $V \ntriangleleft S_4$. – Sammy Black Apr 1 '13 at 6:47
up vote 1 down vote accepted

There's also a lazy way. We can build a map $\varphi: G \to (N/M) \times K$ defined like this: $$ \varphi(nk) = (nM, k) \qquad \textrm{for all }\ n \in N,\ k \in K $$ It is easy to see that $\varphi$ is well defined and is a homomorphism, and $M$ is its kernel, so $M$ is automatically normal in $G$.

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I intuitively thought of something like this but I wasn't quite sure how to do it; making some kind of map that is... =/ – Frudrururu Apr 1 '13 at 7:01
Ok, did this, insightful thank you. – Frudrururu Apr 1 '13 at 7:32

The conjugation route is fine. You have an element $g=nk$ and you want to show that $g^{-1}Mg=M$.

This can easily be done if you first show two things: that $n^{-1}Mn=M$ and that $k^{-1}Mk=M$. One follows from the normality of $M$ in $N$, and the other follows from the fact that elements of $N$ commute with elements of $K$.

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Thank you very much, I will do this. – Frudrururu Apr 1 '13 at 6:53
I'd really appreciate to learn the reason for the downvote. – Dan Shved Apr 3 '13 at 18:43
I am not sure if you are referring to me or not with this but I did not downvote this. Sorry if you weren't referring to me. – Frudrururu Apr 4 '13 at 4:48
That was me. I disapprove of posting multiple answers to the same question. It is unnecessary, and looks like reputation-hunting. – user1729 Apr 4 '13 at 9:46
@user1729 Thanks for the feedback. I just thought it would be more convenient than stacking all the approaches together in one answer (as I usually do). I guess I'll just go back to my old ways then. – Dan Shved Apr 4 '13 at 12:28

Let $N_G(M)= \{ g\in G \mid gMg^{-1}=M \}$ be the normalizer of $M$ in $G$. In fact, $N \subset N_G(M)$ (because $M$ is normal in $N$) and $K \subset N_G(M)$ (because $K \subset C_G(N) \subset C_G(M)$), so $G=NK \subset N_G(M)$ (because $N_G(M)$ is a subgroup of $G$).

Therefore, the key point is that $K \subset C_G(N)$. But for $n \in N$ and $k \in K$, $[n,k]=(nkn^{-1})k^{-1} \in K$ and $[n,k]=n(kn^{-1}k^{-1}) \in N$, hence $[n,k] \in N \cap K$ and $[n,k]=1$.

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