Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a group. A normal subgroup $N$ of $G$ is called ultracharacteristic if for every normal subgroup $U$ of $G$ the condition $$ G/U \cong G/N $$ implies that $U \ge N.$ What are instances of natural conditions on $G/N$ that make $N$ ultracharacteristic? For example, suppose $G/N$ has only inner automorphisms; is then $N$ ultracharacteristic?

share|cite|improve this question
If K is any group, then N = K x {e} is a normal subgroup of G = K x K, with quotient G/N isomorphic to K. But N cannot be ultracharacteristic unless K is trivial, since G/(K x {e}) and G/({e} x K) are isomorphic. So the only condition that references G/N alone and that guarantees N to be ultracharacteristic is "G/N is trivial". – D. Savitt Aug 14 '10 at 16:21
A very neat argument, thanks a lot. – Olod Aug 14 '10 at 16:47
@Savitt: Why don't you add it as answer so people can upvote? – Aryabhata Aug 14 '10 at 18:14
@Moron: To leave some time for a more contentful answer that might tell the asker what question he/she ought to be asking instead.... But since this doesn't seem to be forthcoming, I'll go ahead. – D. Savitt Aug 15 '10 at 0:08
@Savitt: a rephrased/improved question isn't the original one, is it? A lot of people read the question already as it was. Fair's fair. What really interests me is the situation when one has a surjective homomorphism f : G \to GL(n,K) and might hope that the kernel of f is ultracharacteristic. I thought at first the image has something serious to say, but you've demonstrated it's not the case in general. – Olod Aug 15 '10 at 1:05
up vote 5 down vote accepted

If $K$ is any group, then $N = K \times 1$ is a normal subgroup of $G = K \times K$, with quotient $G/N$ isomorphic to $K$. But $N$ cannot be ultracharacteristic unless $K$ is trivial, since $G/(K \times 1)$ and $G/(1 \times K)$ are isomorphic. So the only condition that references $G/N$ alone and that guarantees $N$ to be ultracharacteristic is "$G/N$ is trivial".

share|cite|improve this answer
This also shows there is no condition on N alone. It really depends on the "top" part of the lattice of normal subgroups, and for something like K a simple group, G = K×K, everything is at the top, so you end up looking at all of G anyways, so I don't think you can say too much more that is interesting. – Jack Schmidt Aug 15 '10 at 5:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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