Editing Semidirect product (section)

== Non-examples ==
Of course, no [[simple group]] can be expressed as a semidirect product (because they do not have nontrivial normal subgroups), but there are a few common counterexamples of groups containing a non-trivial normal subgroup that nonetheless cannot be expressed as a semidirect product. Note that although not every group <math>G</math> can be expressed as a split extension of <math>H</math> by <math>A</math>, it turns out that such a group can be embedded into the [[wreath product]] <math>A\wr H</math> by the [[universal embedding theorem]].

=== Z<sub>4</sub> ===
The cyclic group <math>\mathrm{Z}_4</math> is not a simple group since it has a subgroup of order 2, namely <math>\{0,2\} \cong \mathrm{Z}_2</math> is a subgroup and their quotient is <math>\mathrm{Z}_2</math>, so there is an extension <blockquote><math>0 \to \mathrm{Z}_2 \to \mathrm{Z}_4 \to \mathrm{Z}_2 \to 0</math></blockquote>If instead this extension is [[Split extension|split]], then the group <math>G</math> in<blockquote><math>0 \to \mathrm{Z}_2 \to G \to \mathrm{Z}_2 \to 0</math></blockquote>would be isomorphic to <math>\mathrm{Z}_2\times\mathrm{Z}_2</math>.

=== Q<sub>8</sub> ===
The [[Quaternion group|group of the eight quaternions]] <math>\{\pm 1,\pm i,\pm j,\pm k\}</math> where <math>ijk = -1</math> and <math>i^2 = j^2 = k^2 = -1</math>, is another example of a group<ref>{{Cite web|title=abstract algebra - Can every non-simple group $G$ be written as a semidirect product?|url=https://math.stackexchange.com/q/1504422 |access-date=2020-10-29|website=Mathematics Stack Exchange}}</ref> which has non-trivial normal subgroups yet is still not split. For example, the subgroup generated by <math>i</math> is isomorphic to <math>\mathrm{Z}_4</math> and is normal. It also has a subgroup of order <math>2</math> generated by <math>-1</math>. This would mean <math>\mathrm{Q}_8</math> would have to be a split extension in the following ''hypothetical'' exact sequence of groups: <blockquote><math>0 \to \mathrm{Z}_4 \to \mathrm{Q}_8 \to \mathrm{Z}_2 \to 0</math>, </blockquote>but such an exact sequence does not exist. This can be shown by computing the first group cohomology group of <math>\mathrm{Z}_2</math> with coefficients in <math>\mathrm{Z}_4</math>, so <math>H^1(\mathrm{Z}_2,\mathrm{Z}_4) \cong \mathrm{Z}/2</math> and noting the two groups in these extensions are <math>\mathrm{Z}_2\times\mathrm{Z}_4</math> and the dihedral group <math>\mathrm{D}_8</math>. But, as neither of these groups is isomorphic with <math>\mathrm{Q}_8</math>, the quaternion group is not split. This non-existence of isomorphisms can be checked by noting the trivial extension is abelian while <math>\mathrm{Q}_8</math> is non-abelian, and noting the only normal subgroups are <math>\mathrm{Z}_2</math> and <math>\mathrm{Z}_4</math>, but <math>\mathrm{Q}_8</math> has three subgroups isomorphic to <math>\mathrm{Z}_4</math>.