# Can every non-simple group $G$ be written as a semidirect product?

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?

EDIT:

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.

• 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. – Thomas Andrews Oct 30 '15 at 2:14
• @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? – k99731 Oct 30 '15 at 2:22
• @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... – k99731 Oct 30 '15 at 2:55
• – 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.

• 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. – Eli Johnson Oct 17 '17 at 16:32
• @EliJohnson Thanks. Can you edit that in? I cannot do that at the moment! – Pedro Tamaroff Oct 17 '17 at 18:02

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

• Then can a group $G$ that is not simple has a non-trivial semidirect product (where $H$ and $K$ are proper)? – k99731 Oct 30 '15 at 2:29
• @k99731: $\mathbb{Z}_4$ continues to be a counterexample. – Qiaochu Yuan Oct 30 '15 at 2:50
• Yes you are right... I should check that before asking the follow-up question... – k99731 Oct 30 '15 at 2:58
• @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. – Qiaochu Yuan Oct 30 '15 at 3:01