0
$\begingroup$

I’m trying to understand this statement regarding simple groups from Naive Lie Theory by John Stillwell.

The center of any group G is a normal subgroup of G, hence G cannot be simple unless Z(G)={1}.

My question is, what if $Z(G) = G$ (i.e. G is Abelian)? Then the statement is of no value because it simply says $G$ is a normal subgroup of itself, a trivial statement.

If G is a nontrivial Abelian group, can it not be simple? This statement appears to claim that $Z(G) \neq \{1\} \implies G$ is not simple.

It seems to me the textbook is overlooking this important case, when $G$ is nontrivial and Abelian.

$\endgroup$
3
  • 1
    $\begingroup$ Yes, simple abelian groups exist, and they are precisely the finite cyclic groups of prime order. $\endgroup$
    – Ben West
    Aug 30, 2020 at 1:33
  • $\begingroup$ @BW. Great, this is what I expected. Then how can we reconcile Stillwell’s statement? $\endgroup$
    – Raiyan Chowdhury
    Aug 30, 2020 at 1:35
  • 3
    $\begingroup$ You can reconcile Stillwell's statement by adding the hypothesis that $G$ is not abelian. (Maybe this assumption is already elsewhere in the text, I have not read it.) But it is also good to know about the abelian case, which you noticed, to your credit. $\endgroup$
    – Ben West
    Aug 30, 2020 at 1:38

1 Answer 1

Reset to default
2
$\begingroup$

You are right and the book is wrong (or you missed a condition). Any cyclic group of prime order is simple and has non-trivial center.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.