Endomorphism kernel not implies complemented normal

Statement
It is possible to have a group $$G$$ (in fact, we can choose $$G$$ to be a finite group) and a subgroup $$H$$ of $$G$$ such that $$H$$ is an endomorphism kernel in $$G$$ (i.e., there is an endomorphism of $$G$$ with kernel $$H$$) but $$H$$ is not a complemented normal subgroup of $$G$$, i.e., there is no permutable complement to $$H$$ in $$G$$.

Example of cyclic group of prime-square order
Suppose $$p$$ is a prime number and $$G = \mathbb{Z}/p^2\mathbb{Z}$$ is a cyclic group of prime-square order. Let $$H$$ be the unique subgroup of order $$p$$, given by all the elements of order dividing $$p$$, or equivalently, all the elements that are $$p^{th}$$ powers (in additive notation, $$p^{th}$$ multiples).


 * $$H$$ is the kernel of the multiplication by $$p$$ endomorphism and is thus an endomorphism kernel. (In fact, the image of this endomorphism is also $$H$$, so we get $$G/H \cong H$$).
 * $$H$$ has no permutable complement in $$G$$: Every element outside $$H$$ is of order $$p^2$$ and generates $$G$$, so there is no complement to $$H$$ in $$G$$ of order $$p$$.