Group Theory

6 Semidirect Products, Presentations, and Examples

Sometimes we have an exact sequence of groups \(1 \to N \to G \to H \to 1\), that splits, i.e. there is a section of the projection \(G \to H\). In this situation we write \(G \cong N \rtimes _\phi H\) and say that \(G\) is the semidirect product of \(N\) and \(H\). Indeed, \(G\) can be recovered from \(N\) and \(H\) and \(\phi \), the map \(H \to \Aut (N)\) given by conjugation by \(h\). namely, note that \(NH = G\) and \(N\cap H = e\), so we can identify the elements of \(N \times H\) with \(G\) via \((n,h) \to nh\). Now we have \((n,h)(n',h')=nhn'h' = n(hn'h^{-1})hh' = (n\phi (h)(n'),hh')\), so indeed \(N,H,\phi \) are enough data. Conversely any such \(\phi \) gives a group structure on the set \(G \times H\).

Lemma 6.1. If the map \(\phi : H \to \Aut (N)\) changes by an automorphism of \(H\) or by conjugation in \(\Aut (N)\), then the group \(N \rtimes _\phi H\) will not change up to isomorphism.

Proof. The first change obviously doesn’t change anything, and the second corresponds to changing \(N\) by an automorphism. □

The free group construction on a set \(I\) is the group obtained from applying the left adjoint of the forgetful functor \(\Grp \to \Set \). It can be described as words in the letters \(I\) and their inverses, where we can cancel elements with their inverses if they appear next to each other, and multiplication is given by concatenation. We often like to write a group \(G\) as quotients of a free group. A way to do this is as follows: choose generators \(g_\alpha , \alpha \in I\) of \(G\) to determine a surjective map \(F_I \to G\). We can find elements \(r_\beta \) (words in the \(g_\alpha \)) of \(F_I\) that normally generate the kernel, meaning that the smallest normal subgroup that contains them is the kernel. Then we can write \(G = \langle g_\alpha | r_\beta \rangle \), and we can describe \(G\) as words in the \(g_\alpha \), where we are allowed to insert and remove any \(r_\beta \) to the word.

As an example, let’s describe the group \(\ZZ ^2\). It is the abelianization of \(F_2\), the free group on two generators, so we can describe it as \(\langle a,b|[a,b]\rangle \). Then whenever \(ab\) appears in a word, we can replace it with \(ab[a,b]^{-1} = ba\), so indeed \(a\) and \(b\) commute as we might expect.

We say that a group it cyclic if it has \(1\) generator. The free group on one element is \(\ZZ \), so we see all the cyclic groups are \(\ZZ /n\ZZ = \langle a|a^n\rangle \) for \(n \in \ZZ \), which we also denote \(C_n\) when \(n \neq 0\).

As another example, lets consider the symmetry group of a regular \(n\)-gon, a group we will call \(D_n\). It is generated by two elements \(a\), a reflection around an axis, and \(b\), a rotation by \(\frac {2\pi }{n}\). Note that we have the relations \(a^2,b^n,(ab)^2\). Indeed \(D_n = \langle a,b|a^2,b^n,(ab)^2\rangle \), as we can compute the number of elements of each to be \(2n\), by observing that any word can be reduced to \(a^ib^j,0 \leq i <2,0 \leq j < n\). Note that \(b\) generates a normal subgroup of index \(2\), and \(a\) generates a subgroup that doesn’t intersect \(\langle b \rangle \), the subgroup generated by \(b\). Thus \(D_n\) is a semidirect product \(C_n \rtimes _\phi C_2\), where \(\phi \) is the map \(C_2 \to \Aut (C_n)\) sending the nontrivial element to inversion.