# Endomorphism kernel not implies complemented normal

From Groupprops

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) neednotsatisfy the second subgroup property (i.e., complemented normal subgroup)

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## Statement

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

## Proof

### Example of cyclic group of prime-square order

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

Suppose is a prime number and is a cyclic group of prime-square order. Let be the unique subgroup of order , given by all the elements of order dividing , or equivalently, all the elements that are powers (in additive notation, multiples).

- is the kernel of the multiplication by endomorphism and is thus an endomorphism kernel. (In fact, the image of this endomorphism is also , so we get ).
- has no permutable complement in : Every element outside is of order and generates , so there is no complement to in of order .