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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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 .
Example of cyclic group of prime-square order
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 .