Aa.17 Prove that G is cyclic

[karush]

Let G be a group of order 25.
a, Prove that G is cyclic or $g5=e$ for all $g 2 G$.
Generalize to any group of order $p2$ where p is prime.
Let $g\in G$. If $g=e$, then clearly $g^5=e$.
So $g^6=e$. Then $|g|$ divides $25$, i.e., $|g| = 1,5,\textit{ or } 25$.
But $|g|\ne1$ since we assumed $g\ne e$, and $|g|^6=25$
otherwise, G would be cyclic. So $|g|=5, \textit{i.e.,} g^5 = e$.

ok so far anyway



my AA hw

 

[Euge]

If $G$ contains an element of order $25$, then $G$ is cyclic. Otherwise, each non-identity element has order $5$ (by Lagrange's theorem). If $g\in G$ has order $5$, then $g^5 = e$.