Normal not implies image-potentially fully invariant

From Groupprops
Jump to: navigation, search
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., normal subgroup) need not satisfy the second subgroup property (i.e., image-potentially fully invariant subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about normal subgroup|Get more facts about image-potentially fully invariant subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property normal subgroup but not image-potentially fully invariant subgroup|View examples of subgroups satisfying property normal subgroup and image-potentially fully invariant subgroup

Statement

It is possible to have a normal subgroup H of a group G that is not an image-potentially fully invariant subgroup of G, i.e., it is not possible to have a surjective homomorphism \rho:K \to G and a subgroup L of K such that \rho(L) = H.

Related facts

Similar facts

Opposite facts

Proof

Suppose G is the free group of rank two and H is a normal subgroup of G that is not a fully invariant subgroup of G. In other words, there exists an endomorphism \alpha of <mah>G</math> such that \alpha(H) is not contained inside H.

Suppose \rho:K \to G is a surjective homomorphism and L is a subgroup of K such that \rho(L) = H. We show that L is not fully invariant in K.

Suppose N is the kernel of \rho. Since G is a free group, N is a complemented normal subgroup, so there exists a complement G_1 to N in K with an isomorphism \sigma:G_1 \to G such that if \varphi is the retraction with kernel N and image G_1, then \rho = \sigma \circ \varphi. In particular, restricted to G_1, \rho = \sigma.

Now, consider the endomorphism \beta of K defined as \beta = \sigma^{-1} \circ \alpha \circ \rho. Then, we see that \beta(L) = \sigma^{-1}(\alpha(\rho(L)) = \sigma^{-1}(\alpha(H)). Thus, \rho(\beta(L)) = \sigma(\beta(L)) = \alpha(H). But \alpha(H) is not contained in H, so \rho(\beta(L)) is not contained in H, so \beta(L) is not contained in \rho^{-1}(H). In particular, \beta(L) is not contained in L. Hence, L is not fully invariant in K.