# Normal subgroups of infinite groups [duplicate]

Suppose that $$G$$ is an infinite group and $$H\le G$$. The criterion for normal subgroups states that if $$H^g\le H$$ for all $$g\in G$$, then $$H\trianglelefteq G$$ (and thus $$H^g=H$$ for all $$g\in G$$).

I am looking for an example of an infinite group such that $$H^g\le H$$ for some individual $$g\in G$$, but $$H^g\ne H$$.

I know that abelian groups are excluded, since each subgroup of such a group is normal.

Thanks.

• @bof yes, it does work, since $H^g$ is the fixator of $\mathbb{N}_{\geq 2}$ (if we start counting from 1 and not from 0). – frafour Jan 13 at 9:40

There is a construction called HNN-extension that yields plenty of such counterexamples. Let $$G$$ be a group containing two isomorphic subgroups $$H, K$$, and let $$\alpha : H \to K$$ be an isomorphism. The motivating question is: does there exist a supergroup of $$G$$ in which $$H$$ and $$K$$ are not only isomorphic, but conjugate? The answer is yes, and this is called the HNN extension of $$G$$ relative to $$\alpha$$, denoted $$G *_\alpha$$. If the group $$G$$ has a presentation $$\langle S \mid R \rangle$$, you consider a new letter $$t \notin S$$ and define $$G *_\alpha := \langle S, t \mid R, tht^{-1} = \alpha(h) \, \forall \, h \in H \rangle$$. Basically you add a new element whose conjugation realizes the isomorphisms, and you consider the smalles possible such group.
Now in order to get the examples you wanted, you only need a group which is isomorphic to a proper subgroup of itself. For instance you can let $$G = H = \mathbb{Z}$$ and $$K = 2 \mathbb{Z}$$ with $$\alpha(n) = 2n$$. Then $$G = \langle a \mid\rangle$$, so $$G *_\alpha = \langle a, t \mid tat^{-1} = a^2 \rangle$$, and this realizes $$K = H^t < H = G \leq G *_\alpha$$. This group has a name: it is called the Baumslag solitar group, in this case denote $$BS(1, 2)$$.