4

Suppose $1\to K \stackrel{m}{\rightarrow} G \stackrel{f}{\rightarrow} H \to 1$ is short exact sequence of groups.

The followings are equivalent:

$(1)\ G\cong K \times H;$

$(2)$ The sequence right splits (i.e. $\exists$ homomorphism $g:H \to G$ s.t. $f\circ g =$ Id$_H$) and $H\cong N \triangleleft G;$

$(3)\ G$ is semidirect product of $K$ and $H$, and $H$ acts on $K$ trivially.

$(4)$ The sequence left splits (i.e. $\exists$ homomorphism $h:G \to K$ s.t. $h\circ m =$ Id$_K$).

However, if $H$ acts on $K$ nontrivially, $G$ may also be direct product.

e.g. let $G$ be a nonabelian group with $h \in G\backslash Z(G)$.

for $1\to G \to G\times \Bbb Z \to \Bbb Z \to 1$, splitting map $g: \Bbb Z \to G \times \Bbb Z, 1 \mapsto (h,1).$

$ \phi: \Bbb Z \to \text{Aut} G,\ \phi(1)(g,1)=(h,1)(g,1)(h,1)^{-1}=(hgh^{-1},1).$

Since $h \not \in Z(G)$, this action is nontrivial.

So under what condition, semidirect product of groups is isomorphic to their direct products?

And more generally, when is $A\rtimes_{\phi_1} B \cong A\rtimes_{\phi_2} B$?

Andrews
  • 3,953
  • 4
    What exactly are you asking? As far as I can see everything you have written is correct, and there is no contradiction. It is possible for a direct product $K \times H$ to be isomorphic to a semidirect product $K \rtimes H$ with $H$ acting nontrivially on $K$. A smaller example is $K=S_3$, $H=C_2$, where the nontrivial action of $H$ on $K$ is an inner automorphism. – Derek Holt Feb 09 '19 at 15:23
  • "when is $A\rtimes_{\phi_1} B \cong A\rtimes_{\phi_2} B$?" AFAIK there is no general rule and these have to be considered case by case. You may also want to read this: https://math.stackexchange.com/questions/527800/when-are-two-semidirect-products-isomorphic – freakish Feb 14 '19 at 12:34

1 Answers1

2

There're a number of questions and answers for this problem in MSE (as you can see on the right), and I'd like to make a summary.

For groups $N,H$, group homomorphism $\varphi,\phi:H\to \text{Aut }N$ represents an action of $H$ on $N$.

$(1)$ If $\exists f\in\text{Aut}(H),g\in \text{Aut}(N)$ s.t. $\varphi\circ f=c_g \circ \phi$, where $c_g \circ\phi(h)=g\phi(h)g^{-1}$,then

$N\rtimes_\phi H\cong N \rtimes_\varphi H$ via $\psi:(n,h)\mapsto(g(n),f(h))$.

$(2)$ If im$\varphi\subset \text{Inn }N$, then $G=NC_G(N).$ And since $C_G(N)\triangleleft G$, if in addition $Z(N)=1$

i.e. $N\cap C_G(N)=1,$ then $G\cong N\times C_G(N)$ and $G/N\cong C_G(N)$.

From $G/N \cong H, H \cong C_G(N)$ and we have $N\rtimes_\varphi H\cong G\cong N \times C_G(N)\cong N \times H.$

Andrews
  • 3,953