# Showing that $G/N \cong A \times (B / N)$

Let $$G$$ be the semi-direct product of a group $$A$$ by a group $$B$$ where $$B$$ is normal in $$G$$. Furthermore, let $$N = [A,B]$$ denote the subgroup of $$G$$ generated by all commutators $$[a, b] = aba^{-1}b^{-1}$$, where $$a \in A$$ and $$b \in B.$$

I am trying to show that $$G/N \cong A \times (B / N)$$.

Here is my thought process: First I show that $$N$$ is normal in $$G$$. ( This is straight forward and has been done). Secondly I define a mapping $$\varphi: G \rightarrow A \times (B / N)$$ $$g= ab \mapsto (a, b[a,b])$$ I will need to show that $$\varphi$$ is a surjective homomorphism whose kernel is $$G'$$. Then by the first isomorphism theorem I can conclude that $$G/N \cong A \times (B / N)$$.

My problem is showing that $$\varphi$$ is a homomorphism. That is to say for any $$g_1, g_2 \in G$$ such that $$g_1 = ab$$ , $$g_2 = cd$$ where $$a,c \in A$$, and $$b,d \in B$$, we have that $$\varphi(g_1 \cdot g_2) = \varphi(g_1) \cdot \varphi(g_2)$$.

Is my mapping sufficient? Is my thought process correct? Any help would be appreciated.

• Please don't make subject lines that consist entirely of MathJax: it prevents certain navigation shortcuts. Commented Jul 5, 2021 at 15:22
• The notation $G'$ is unfortunate: this is usually the notation for the derived subgroup of $G$, which is defined as $[G,G]$. Also, please clarify: when you say "semidirect product of $A$ by $B$", I would interpret that as $A$ being normal and $B$ acting on $A$. But if that is the case, then $G'$ would be a subgroup of $A$, and you seem to be thinking it is a subgroup of $B$. Which one is the normal subgroup, and which one is acting on it? Commented Jul 5, 2021 at 15:23
• Please make that clear in the post. Commented Jul 5, 2021 at 15:37
• Your map makes no sense. $b[a,b]$ is an element of $B$, not an element of $B/N$. You would need to send $b$ to $bN$. Commented Jul 5, 2021 at 15:51
• That, at least, makes sense. Commented Jul 5, 2021 at 16:22

More generally, suppose that $$G=K\rtimes H$$ (here $$K$$ is the normal subgroup, $$H$$ acts on it), and that $$N$$ is a subgroup of $$K$$ that is normal in $$K$$, and that is $$H$$-invariant; that is, for all $$n\in N$$ and $$h\in H$$, $$hnh^{-1}\in N$$. We want to show that $$N$$ is normal in $$G$$, and that $$G/N\cong (K/N)\rtimes H$$, with the action of $$H$$ on $$K/N$$ being the one induced by the action of $$H$$ in $$K$$.

Though I am thinking of $$G$$ as an internal semidirect product, I will use the ordered pair notation: so elements of $$G$$ are written uniquely as $$(h,k)$$ with $$h\in H$$, $$k\in K$$, and the product is given by $$(h,k)*(h',k') = (hh',k^{h'}k') = (hh',(h')^{-1}kh'kk').$$

Indeed, since $$G$$ is generated by $$K$$ and $$H$$, and $$N$$ is invariant under conjugation by elements of both $$K$$ and $$H$$, it follows that $$N\triangleleft G$$. Note also that the action of $$H$$ on $$K$$ induces an action of $$H$$ on $$K/N$$ by conjugation, because $$N$$ is mapped to itself.

To show that $$G/N\cong (K/N)\rtimes H$$, we take the map defined (as you kind of almost did) by $$\varphi\colon G\to (K/N)\rtimes H$$ by $$\varphi(h,k) = (h,kN)$$.

This is a homomorphism: given $$(h_1,k_1)$$, $$(h_2,k_2)$$ in $$G$$, we have \begin{align*} \varphi(h_1,k_1)\varphi(h_2,k_2) &= (h_1,k_1N)(h_2,k_2N)\\ &= (h_1h_2, (k_1N)^{h_2}k_2N) \\ &= (h_1h_2,(k_1^{h_2}Nk_2N)\\ &= (h_1h_2,k_1^{h_2}k_2N);\\ &=\varphi(h_1h_2,k_1^{h_2}k_2)\\ &= \varphi\bigl( (h_1,k_1)(h_2,k_2)\bigr). \end{align*} The kernel of $$\varphi$$ consists of the $$(h,k)\in G$$ such that $$\varphi(h,k)=(h,kN)=(e,eN)$$. Thus, we must have $$h=e$$ and $$k\in N$$. Conversely, all such elements are in the kernel.

Finally, $$\varphi$$ is surjective, since the element $$(h,kN)$$ of $$(K/N)\rtimes H$$ is the image of $$(h,k)\in G$$. Thus, we have that $$\frac{G}{N} \cong \frac{K}{N}\rtimes H.$$

Now, in your specific situation, $$N=[K,H]$$. This is a subgroup of $$K$$, since the generators all lie in $$K$$: $$[k,h] = k(hk^{-1}h^{-1})\in K$$.

It is invariant under the action of $$H$$, since given a generator $$[k,h]$$ and $$y\in H$$, we have $$y[k,h]y^{-1}=[yky^{-1},yhy^{-1}]\in [K,H]$$ (since $$yky^{-1}\in K$$).

It is normal in $$K$$ (see, e.g., here, though that uses a different commutator convention), since given $$x\in K$$ and a generator $$[k,h]=khk^{-1}h^{-1}$$ of $$[K,H]$$, we have \begin{align*} x[k,h]x^{-1} &= xkhk^{-1}h^{-1}x^{-1}\\ &= (xkhk^{-1})(x^{-1}h^{-1}hx)h^{-1}x^{-1}\\ &= \bigl((xk)h(xk)^{-1}h^{-1}\bigr) \bigl(hxh^{-1}x^{-1}\bigr)\\ &= [xk,h][h,x]\\ &= [xk,h][x,h]^{-1}. \end{align*} This element lies in $$[K,H]$$, since it is the product of a generator and the inverse of a generator of $$[K,H]$$.

Thus, in your situation we have $$G/[K,H]\cong (K/[K,H])\rtimes H$$.

It only remains to show that this semidirect product is actually a direct product. For that, we just need to show that the action of $$H$$ on $$K/[K,H]$$ is trivial. Indeed, letting $$N=[K,H]$$, let $$h\in H$$ and $$k\in K$$. Then $$(kN)^h = h^{-1}khN = kk^{-1}h^{-1}khN = k([k^{-1},h^{-1}])N = kN,$$ since $$[k^{-1},h^{-1}]\in N$$. Thus, this is a semidirect product with a trivial action, and so isomorphic to the direct product, as desired.