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.


  • $\begingroup$ @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). $\endgroup$ – 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 $BS(1, 2)$.

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