# Characteristicity does not satisfy image condition

This article gives the statement, and possibly proof, of a subgroup property (i.e., characteristic subgroup) not satisfying a subgroup metaproperty (i.e., image condition).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties

## Statement

### Property-theoretic statement

The subgroup property of being a characteristic subgroup satisfies the subgroup metaproperty called the image condition.

### Statement with symbols

If $H$ is a characteristic subgroup of $G$ and $f:G \to K$ is a surjective homomorphism, then $f(H)$ is not necessarily a characteristic subgroup of $K$.

## Partial truth

Although the image of every characteristic subgroup under a surjective homomorphism need not be characteristic, there are special situations where we can guarantee this property. A characteristic subgroup whose image under every surjective homomorphism is characteristic is termed an image-closed characteristic subgroup.

It turns out that every verbal subgroup, and in particular, every member of the lower central series and derived series, as well as the agemo subgroups, are image-closed characteristic. In fact, their images are verbal subgroup with the same defining words.

On the other hand, the center need not satisfy the property that its image is always characteristic. For full proof, refer: Center not is image-closed characteristic

## Proof

We present here two counterexamples of groups of order eight.

### Example of an Abelian group of order eight

Let $G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$, and $H$ be the subgroup $\{ (0,0), (0,2), (1,0), (1,2) \}$. This is $\Omega_1(G)$ (the first omega subgroup) -- the set of elements of order two in $G$. Thus, $H$ is a characteristic subgroup of $G$.

Let $N = \{ (0,0), (0,2) \}$, and $f:G \to G/N$ be the quotient map. Then, $f(G) = G/N$ is isomorphic to the Klein-four group, and $f(H) = H/N$ is a subgroup of order two in it, which is clearly not characteristic in $f(G) = G/N$.

### Example of the dihedral group

Consider the dihedral group of order eight with an element $a$ of order four, playing the role of the rotation, and an element $x$ of order two, playing the role of the reflection. Let $H$ be the cyclic subgroup of order four generated by $a$. Then, $H$ is a characteristic subgroup of $G$.

Consider the quotient map $f:G \to G/N$ where $N = \{ e, a^2\}$. The group $G/N$ is isomorphic to a Klein-four group, and the image $f(H) = H/N$ is a subgroup of order two in that Klein-four group. Clearly, $f(H)$ is not characteristic inside $K$.