My question is:
What are the implications if the order of the factor group $G/Z(G)$ is equal to the order of the group $G$?
I know that $|G/Z(G)|=|G|$ if $|Z(G)|=1$ meaning the center is trivial.
Since $G/Z(G)$ is the set of all left cosets then if for example say $G=\{a,b,c,d,e\}$ then $G/Z(G)=\{\{a\},\{b\},\{c\},\{d\},\{e\}\}$
What can I say about $G$ based from these? Can I say that $G$=$Z(G)$?
This question arose when I tried to answer the following question:
Every group of order 203 with a normal subgroup of order 7 is abelian
Here is my solution to this question:
Here is a try at a solution but I'm having trouble showing something.
Consider the center of G $Z(G)$. We know that $Z(G)\leq G$
By Lagrange's theorem $|Z(G)|$ must divide $|G|$.
So therefore the possible order of $Z(G)$ are $1, 7, 29, 203$.
If $|Z(G)|=203$ then $Z(G)=G$ and $H\leq G=Z(G)$ and $G$ is abelian.
Now $|Z(G)|\neq29$ because if $|Z(G)|=29$ then the order of the factor group $G/Z(G)$ is $|G|/|Z(G)|$ which is equal to $203/29 = 7$.
So order of $G/Z(G)$ is prime $\implies$ $G/Z(G)$ is cyclic $\implies$ $G$ is abelian $\implies G=Z(G)$ which is a contradiction since $|G|=203$.
The same argument also shows that $|Z(G)|\neq7$.
I wanted to show that $|Z(G)|\neq1$.
So I'm asking what are the implications of having a trivial center.
Thank you.
