Endomorphism kernel not implies complemented normal

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., endomorphism kernel) need not satisfy the second subgroup property (i.e., complemented normal subgroup)
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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 $G$.

Proof

Example of cyclic group of prime-square order

Further information: cyclic group of prime-square order, cyclic group:Z4

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 $p$.