Statements about group-order-centralizer

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In summary, you have shown that if $N\leq Z(G)$ and $G/N$ is cyclic, then $G$ is abelian, and if $G$ is a non-abelian group of order $p^3$ where $p$ is a prime, then $|Z(G)|=p$.
  • #1
mathmari
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Hey! :giggle:

a) Let $N\leq Z(G)$ where $Z(G)$ is the centralizer of a group $G$. Show that if $G/N$ is cyclic then $G$ is abelian.
b) Let $G$ be a non-abelian group of order $p^3$ where $p$ a prime. Show that $|Z(G)|=p$. I have done the following :

a) Since $N\leq Z(G)$ we have that $$G/N \ \text{ cyclic } \ \iff G/Z(G) \ \text{ cyclic } \ $$
It holds that $G$ abelian $\iff$ $\ Z(G)=G$.
So we have to show that the only element of $G/Z(G)$ is the identity class $Z(G)$.
Let $G/Z(G)=\langle gZ(G)\rangle$ and let $a\in G$.
Then there is $i$ such that $aZ(G)=(gZ(G))^i=g^iZ(G)$.
So $a=g^iz$ for some $z\in Z(G)$.
Since $g^i, z\in C(g)$ then $a\in C(g)$.
Since $a\in G$ arbitrary, we have that each element of $G$ has the commutative property with $g$, i.e. $z\in Z(G)$.
So $gZ(G)=Z(G)$ is the only element of $G/Z(G)$.

b) We have that $Z(G)\leq G$. From Lagrange Theorem we have that $|Z(G)|\mid |G|$.
Since $|G|=p^3$ the possible orders of $Z(G)$ are $1, p, p^2, p^3$.
We have that $|Z(G)|\neq p^3$ because in other case we would have $Z(G)=G$, but $G$ is non-abelian (since $G$ abelian $\iff$ $\ Z(G)=G$).
We have that $|Z(G)|\neq p^2$ because in other case we would have $|G/Z(G)|=\frac{|G|}{|Z(G)}|=\frac{p^3}{p^2}=p$ and then $|G/Z(G)|=p$ prime and so $G/Z(G)$ cyclic and so $G$ abelian. A contradiction, since $G$ is non abelian.
We have that $|Z(G)|\leq 1$ because $G$ is a $p$-group and so it cannot have a trivial center.
Therefore the only possibe order that remains is $|Z(G)|=p$. Is everything correct and complete? :unsure:
 
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  • #2


Yes, everything appears to be correct and complete. Your proof for part a) is clear and well-explained, and your reasoning for part b) is also sound. Great job!
 

FAQ: Statements about group-order-centralizer

1. What is the definition of group-order-centralizer?

Group-order-centralizer is a mathematical concept that refers to the set of elements in a group that commute with every element in a given subgroup of that group.

2. How is group-order-centralizer related to group theory?

Group-order-centralizer is a fundamental concept in group theory, as it helps to understand the structure and properties of groups. It is used to classify groups and determine their subgroups.

3. How is group-order-centralizer different from a normal subgroup?

While group-order-centralizer is a subgroup that commutes with every element in a given subgroup, a normal subgroup is a subgroup that is invariant under conjugation by any element in the group.

4. Can a group have more than one group-order-centralizer?

Yes, a group can have multiple group-order-centralizers, as long as there are different subgroups within the group that have elements that commute with every element in that subgroup.

5. Why is group-order-centralizer important in group theory?

Group-order-centralizer helps to classify groups and understand their structure and properties. It also has applications in other areas of mathematics, such as algebraic geometry and number theory.

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