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Let $G$ be a finite group and $D \subset G$ be a Hadamard difference set in $G$. If $D^{-1}=\{d^{-1}| d \in D\}$ is also a difference set in $G$ which is equivalent to $D$, then what can we say about the group $G$? Does this group have some properties?

Comments: A $(n,k,\mu)$-difference set in a finite group $G$ of size $n$ is a subset $D \subseteq G$ of size $k$ such that every nonidentity element of $G$ can be expressed exactly $\mu$-times as $d_{1}d_{2}^{-1}$ of elements of $D$.

Two difference sets $D_{1}$ and $D_{2}$ in a group $G$ are equivalent if there is an automorphism $\Psi \in Aut(G)$ such that $\Psi(D_{1})=\{\Psi(d)|d \in D_{1}\}=gD_{2}$ for some $g \in G$.

$(n,k,\mu)=(4u^{2},2u^{2}-u,u^{2}-u)$ for some integer number $u$ is related to Hadamard difference sets.

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    $\begingroup$ It's not so common background; the question would benefit if you copy the definition of difference set, and also of being equivalent. $\endgroup$
    – YCor
    Commented Mar 26, 2019 at 15:58
  • $\begingroup$ A $(n,k,\mu)$-difference set in a finite group $G$ of size $n$ is a subset $D \subseteq G$ of size $k$ such that every nonidentity element of $G$ can be expressed exactly $\mu$-times as $d_{1}d_{2}^{-1}$ of elements of $D$. $\endgroup$
    – Mojtaba Jazaeri
    Commented Mar 26, 2019 at 17:27
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    $\begingroup$ Yes, please edit your question to add this, and also please add what it means for a difference set to be Hadamard, and what it means for two difference sets to be equivalent. Thank you :-) $\endgroup$
    – Franka Waaldijk
    Commented Mar 26, 2019 at 17:45
  • $\begingroup$ The group $G$ is not abelian in general. $\endgroup$
    – Mojtaba Jazaeri
    Commented Mar 26, 2019 at 17:59
  • $\begingroup$ yes sorry, i deleted my comment but not in time... $\endgroup$
    – Franka Waaldijk
    Commented Mar 26, 2019 at 18:00

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