# Isomorphic to inner automorphism group not implies centerless

This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., group isomorphic to its inner automorphism group) need not satisfy the second group property (i.e., centerless group)
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## Statement

There can exist a group $G$ such that the Center (?) of $G$ is nontrivial, but $G$ is isomorphic to the quotient group $G/Z(G)$ (which is also the Inner automorphism group (?) of $G$).

## Proof

### Example of a generalized dihedral group

Further information: generalized dihedral group of 2-quasicyclic group

Let $H$ be the group of all $(2^n)^{th}$ roots of unity among the complex numbers, under multiplication, for all $n$. Let $G$ be the semidirect product of $H$ with a cyclic group of order two acting by the inverse map. In other words, $G$ is the generalized dihedral group corresponding to the abelian group $H$. Then, we have:

• The center $Z(G)$ of $G$ is a group of order two inside $H$, namely, the elements $\pm 1$: Clearly, no element outside $H$ is in the center, because any element outside $H$ acts on $H$ by the inverse map under conjugation. This leaves elements inside $H$. Any central element of $H$ must be fixed by conjugation by elements outside $H$, namely, the inverse map. But there are only two elements of $H$ that are fixed by the inverse map, namely $\pm 1$.
• The quotient $G/Z(G)$ is isomorphic to $G$: Note that $H/Z(G)$ is isomorphic to $H$, via the map sending $x$ to $x^2$ (the map is well-defined because both elements in the coset of $x$, namely $x$ and $-x$, get sent to the same element: $x^2$). Further, the inverse map commutes with this map, so it extends to an isomorphism between $G/Z(G)$ and $G$.