MHB Is Every Group of Order 25 Cyclic?

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Cyclic
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
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
 
Physics news on Phys.org
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 $g^5 = e$.
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top