Given a group \(G\) with identity element \(e\), a subgroup \(H\), and a normal subgroup \(N \trianglelefteq G\); then we say that \(G\) is the semi-direct product of \(N\) and \(H\) (written \(G=N \rtimes H\)) if \(G\) is the product of subgroups, \(G = NH\) where the subgroups have trivial intersection \(N \cap H= \{e\}\).

Semi-direct products of groups provide examples of non abelian groups. For example the dihedral group \(D_{2n}\) with \(2n\) elements is isomorphic to a semidirect product of the cyclic groups \(\mathbb Z_n\) and \(\mathbb Z_2\). \(D_{2n}\) is the group of isometries preserving a regular polygon \(X\) with \(n\) edges.

Let’see that the converse is not true and present a group that is not a semi-direct product.

The Hamilton’s quaternions group is not a semi-direct product

The Hamilton’s quaternions group \(\mathbb H_8\) is the group consisting of the symbols \(\pm 1, \pm i, \pm j, \pm k\) where\[
-1 = i^2 =j^2 = k^2 \text{ and } ij = k = -ji,jk = i = -kj, ki = j = -ik.\] One can prove that \(\mathbb H_8\) endowed with the product operation above is indeed a group having \(8\) elements where \(1\) is the identity element.

\(\mathbb H_8\) is not abelian as \(ij = k \neq -k = ji\).

Let’s prove that \(\mathbb H_8\) is not the semi-direct product of two subgroups. If that was the case, there would exist a normal subgroup \(N\) and a subgroup \(H\) such that \(G=N \rtimes H\).

• If \(\vert N \vert = 4\) then \(H = \{1,h\}\) where \(h\) is an element of order \(2\) in \(\mathbb H_8\). Therefore \(h=-1\) which is the only element of order \(2\). But \(-1 \in N\) as \(-1\) is the square of all elements in \(\mathbb H_8 \setminus \{\pm 1\}\). We get the contradiction \(N \cap H \neq \{1\}\).
• If \(\vert N \vert = 2\) then \(\vert H \vert = 4\) and \(H\) is also normal in \(G\). Noting \(N=\{1,n\}\) we have for \(h \in H\) \(h^{-1}nh=n\) and therefore \(nh=hn\). This proves that the product \(G=NH\) is direct. Also \(N\) is abelian as a cyclic group of order \(2\). \(H\) is also cyclic as all groups of order \(p^2\) with \(p\) prime are abelian. Finally \(G\) would be abelian, again a contradiction.

We can conclude that \(G\) is not a semi-direct product.

Let’s \(G\) be a group. A characteristic subgroup is a subgroup \(H \subseteq G\) that is mapped to itself by every automorphism of \(G\).

An inner automorphism is an automorphism \(\varphi \in \mathrm{Aut}(G)\) defined by a formula \(\varphi : x \mapsto a^{-1}xa\) where \(a\) is an element of \(G\). An automorphism of a group which is not inner is called an outer automorphism. And a subgroup \(H \subseteq G\) that is mapped to itself by every inner automorphism of \(G\) is called a normal subgroup.

Obviously a characteristic subgroup is a normal subgroup. The converse is not true as we’ll see below.

Example of a direct product

Let \(K\) be a nontrivial group. Then consider the group \(G = K \times K\). The subgroups \(K_1=\{e\} \times K\) and \(K_2=K \times \{e\} \) are both normal in \(G\) as for \((e, k) \in K_1\) and \((a,b) \in G\) we have
\[(a,b)^{-1} (e,x) (a,b) = (a^{-1},b^{-1}) (e,x) (a,b) = (e,b^{-1}xb) \in K_1\] and \(b^{-1}K_1 b = K_1\). Similar relations hold for \(K_2\). As \(K\) is supposed to be nontrivial, we have \(K_1 \neq K_2\).

The exchange automorphism \(\psi : (x,y) \mapsto (y,x)\) exchanges the subgroup \(K_1\) and \(K_2\). Thus, neither \(K_1\) nor \(K_2\) is invariant under all the automorphisms, so neither is characteristic. Therefore, \(K_1\) and \(K_2\) are both normal subgroups of \(G\) that are not characteristic.

When \(K = \mathbb Z_2\) is the cyclic group of order two, \(G = \mathbb Z_2 \times \mathbb Z_2\) is the Klein four-group. In particular, this gives a counterexample where the ambient group is an abelian group.

Example on the additive group \(\mathbb Q\)

Consider the additive group \((\mathbb Q,+)\) of rational numbers. The map \(\varphi : x \mapsto x/2\) is an automorphism. As \((\mathbb Q,+)\) is abelian, all subgroups are normal. However, the subgroup \(\mathbb Z\) is not sent into itself by \(\varphi\) as \(\varphi(1) = 1/ 2 \notin \mathbb Z\). Hence \(\mathbb Z\) is not a characteristic subgroup.

I here provide a simple example of a group whose set of commutators is not a subgroup. The counterexample is due to P.J. Cassidy (1979).

Description of the group \(G\)

Let \(k[x,y]\) denote the ring of all polynomials in two variables over a field \(k\), and let \(k[x]\) and \(k[y]\) denote the subrings of all polynomials in \(x\) and in \(y\) respectively. \(G\) is the set of all upper unitriangular matrices of the form
\[A=\left(\begin{array}{ccc}
1 & f(x) & h(x,y) \\
0 & 1 & g(y) \\
0 & 0 & 1 \end{array}\right)\] where \(f(x) \in k[x]\), \(g(y) \in k[y]\), and \(h(x,y) \in k[x,y]\). The matrix \(A\) will also be denoted \((f,g,h)\).
Let’s verify that \(G\) is a group. The products of two elements \((f,g,h)\) and \((f^\prime,g^\prime,h^\prime)\) is
\[\left(\begin{array}{ccc}
1 & f(x) & h(x,y) \\
0 & 1 & g(y) \\
0 & 0 & 1 \end{array}\right)
\left(\begin{array}{ccc}
1 & f^\prime(x) & h^\prime(x,y) \\
0 & 1 & g^\prime(y) \\
0 & 0 & 1 \end{array}\right)\]
\[=\left(\begin{array}{ccc}
1 & f(x)+f^\prime(x) & h(x,y)+h^\prime(x,y)+f(x)g^\prime(y) \\
0 & 1 & g(y)+g^\prime(y) \\
0 & 0 & 1 \end{array}\right)\] which is an element of \(G\). We also have:
\[\left(\begin{array}{ccc}
1 & f(x) & h(x,y) \\
0 & 1 & g(y) \\
0 & 0 & 1 \end{array}\right)^{-1} =
\left(\begin{array}{ccc}
1 & -f(x) & f(x)g(y) – h(x,y) \\
0 & 1 & -g(y) \\
0 & 0 & 1 \end{array}\right)\] proving that the inverse of an element of \(G\) is also an element of \(G\). Continue reading The set of all commutators in a group need not be a subgroup →