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 →