# Normal not implies image-potentially fully invariant

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)
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## 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 $\rho(L) = H$.

## 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[/itex] 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 $K$.