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

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 $$H$$.

Related facts

 * Exponent of extension group is a multiple of lcm of exponents of normal subgroup and quotient group
 * Exponent of extension group divides product of exponents of normal subgroup and quotient group
 * Exponent of semidirect product may be strictly less than product of exponents

Example of dihedral group
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 $$H$$.