English (United Kingdom)

https://curated-unify.zendy.io/wp-json/zendy-region/v1/featured_content/oa?rat=en

https://curated-unify.zendy.io/wp-json/zendy-region/v1/highlighted_journal/

Global: Zendy Plus annual

Presents the access of premium content as premium feature

Premium Content

Presents the keyphrase highlighting as premium feature

Keyphrase Highlighting

Presents the summarisation as premium feature

Summarisation

Insights

Presents the pdf analysis as premium feature

PDF Analysis

Presents the zaia usage as premium feature

ZAIA

Global: Zendy Plus monthly

Global: Zendy Tools annual

Global: Zendy Tools monthly

Global: Zendy Free Plan

Let $o(G)$ be the average order of a finite group $G$. M. Herzog, P.Longobardi and M. Maj \cite{7} showed that if $G$ is non-solvable and$o(G)=o(A_5)$, then $G\cong A_5$. In this note, we prove that the equality$o(G)=o(A_5)$ does not hold for any finite solvable group $G$. Consequently, upto isomorphism, $A_5$ is determined by its average order.

A characterization of $A_5$ by its average order