# Sufficient conditions for ultracharactericity

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?

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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
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".