Exponent of semidirect product may be strictly greater than lcm of exponents

Statement

In terms of internal semidirect products

It is possible to have a finite group $G$, a complemented normal subgroup $N$ of $G$ with a complement $H$ (so $G$ is an internal semidirect product of $N$ and $H$) such that the exponent of $G$ is strictly greater than the lcm of the exponents of $N$ and $H$.

Proof

Example of dihedral group

Further information: dihedral group:D8, subgroup structure of dihedral group:D8

Consider $G$ to be the dihedral group:D8: $G = \langle a,x \mid a^4 = x^2 = e, xax = a^{-1} \rangle$

This has eight elements: $\! G = \{ e,a,a^2,a^3,x,ax,a^2x,a^3x\}$

Suppose $N$ is one of the Klein four-subgroups of dihedral group:D8: $\! N = \{ e, a^2, x, a^2x \}$

and take $H$ as one of the non-normal subgroups of dihedral group:D8 that is not contained in $N$: $\! H = \{ e, ax \}$

We note that:

• The exponent of $G$ is 4.
• The exponent of $N$ is 2.
• The exponent of $H$ is 2.

Thus, the exponent of $G$ is strictly greater than the lcm of the exponents of $N$ and $H$.