When people want to classify a group with certain (small) order, they seem to find a normal subgroup $H$ and a subgroup $K$, and then consider how many distinct $H \rtimes_\varphi K$ are there.

My question is: For any group $G$, does it always exist normal subgroup $H$, subgroup $K$ and $\phi : K \rightarrow Aut(H)$ s.t. $G\cong H \rtimes_\varphi K $?

Sorry if this question is too basic...

Thanks the comment and the answers. Then if the $G$ is not simple, can we have can $G\cong H\rtimes_\varphi K$ for proper subgroup $H$, $K$, with $H$ normal?


If we have a group $G$, that is not simple (then $G$ has at least one normal subgroup), are there always a non-trivial normal subgroup $H$ and a proper subgroup $K$ s.t. $G \cong H \rtimes K$. (Note that if $G$ is simple this is clearly not possible).

But the quaternion group and $Z_4$ given in answers have shown that we cannot always have it.

  • 1
    $\begingroup$ Presumably, you want non-trivial cases. In any event, no, there are groups called simple groups with no non-trivial normal subgroups. And there are groups which have normal subgroups but which cannot be represented as a semi-direct product. $\endgroup$ – Thomas Andrews Oct 30 '15 at 2:14
  • $\begingroup$ @ThomasAndrews What if $G$ has at least one non-trivial normal subgroup, can $G \cong H \rtimes_\varphi K$ for proper subgroup $H$, $K$, with $H$ normal? $\endgroup$ – k99731 Oct 30 '15 at 2:22
  • $\begingroup$ @PaulPlummer What I try to ask is, if we have a group $G$ that is not simple, are there always a non-trivial normal subgroup $H$ and a proper subgroup $K$ s.t. $G \cong H \rtimes K$. (Note that if $G$ is simple this is clearly not possible). But quaternion group and $Z_4$ seem to be good counter-example... $\endgroup$ – k99731 Oct 30 '15 at 2:55
  • $\begingroup$ See also math.stackexchange.com/questions/127090/…. $\endgroup$ – Noah Schweber Oct 30 '15 at 4:32

A nice and minimal nonabelian example is the quaternion group $Q$. We can define $Q$ as the group of elements $\pm 1,\pm i,\pm j,\pm k$ such that $ijk=-1$ and $i^2=j^2=k^2=-1$. This group has order $8$, so if it has any chance of being a (nontrivial) semidirect product, it has to be one of groups of order $2$ or $4$. Now $Q$ contains only one element of order $2$; namely $-1$. But an iterated semidirect product of $C_2$s will contain more than one such element, since if $x$ has order $2$ in the left factor, then $(x,0)$ and $(e,1)$ both have order $2$ in the semidirect product. The same is true for the semidirect product of $C_2$ and $C_4$ (in both cases the group obtained is $D_8$).

Note that $Q$ is not simple, in fact every subgroup of $Q$ is normal. Another way to see $Q$ is not a semidirect product is to note every pair of subgroups have nontrivial intersection, in fact the intersection of all nontrivial subgroups of $Q$ is the center $Z(Q)=\{-1,1\}$, but in a semidirect product there always are subgroups that have trivial intersection. Yet another way to see this is that since every subgroup is normal, $Q$ would have to be abelian if it were a semidirect product, and it isn't.

Note, however, that $Q$ is a quotient of a semidirect product, namely we can obtain $Q$ has a quotient of $H=C_4\rtimes C_4$ since $Q$ can be presented as $$\langle a,b\mid a^4,b^4,aba^{-1}=b^{-1},a^2b^2\rangle $$

so it suffices we quotient $H$ by the subgroup generated by $a^2b^2$ where $a,b$ are generators of $C_4$ in each summand.

  • $\begingroup$ A small correction: an iterated semidirect product of $C_2$s is not necessarily abelian. Take for example $(C_2\times C_2)\rtimes C_2$, where the automorphism associated to $\phi(1)$ permutes two generators. If $a$ is a permuted generator, $(a,0)\cdot(a,1)\not = (a,1)\cdot (a,0)$. The resulting group has five elements of order 2, so $Q$ is still a counterexample. $\endgroup$ – Eli Johnson Oct 17 '17 at 16:32
  • $\begingroup$ @EliJohnson Thanks. Can you edit that in? I cannot do that at the moment! $\endgroup$ – Pedro Tamaroff Oct 17 '17 at 18:02

No. The smallest example is $\mathbb{Z}_4$. See also simple group and group extension.

  • $\begingroup$ Then can a group $G$ that is not simple has a non-trivial semidirect product (where $H$ and $K$ are proper)? $\endgroup$ – k99731 Oct 30 '15 at 2:29
  • 1
    $\begingroup$ @k99731: $\mathbb{Z}_4$ continues to be a counterexample. $\endgroup$ – Qiaochu Yuan Oct 30 '15 at 2:50
  • $\begingroup$ Yes you are right... I should check that before asking the follow-up question... $\endgroup$ – k99731 Oct 30 '15 at 2:58
  • $\begingroup$ @k99731: the best thing you can hope for is that if $G$ isn't simple, then at least by definition it has some normal subgroup $H$, so it fits into a short exact sequence $1 \to H \to G \to G/H \to 1$. If you know what $H$ and $G/H$ are, you can try to figure out what $G$ is by solving the group extension problem. $\endgroup$ – Qiaochu Yuan Oct 30 '15 at 3:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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