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

This question already has an answer here:

Let $G$ be a group, $H<G$ a subgroup and $g$ an element of $G$. Let $\lambda_g$ denote the inner automorphism which maps $x$ to $gxg^{-1}$. I wonder if $H$ can be mapped to a proper subgroup of itself, i.e. $\lambda_g(H)\subset H$.

I tried to approach this problem topologically. Since every group is the fundamental group of a connected CW-complex of dimension 2, let $(X,x_0)$ be such a space for $G$. Since $X$ is (locally) path-connected and semi-locally simply-connected, there exists a (locally) path-connected covering space $(\widetilde X,\widetilde x_0)$, such that $p_*(\pi_1(\widetilde X,\widetilde x_0))=H$. The element $g$ corresponds to $[\gamma]\in\pi_1(X,x_0))$, and its lift at $\widetilde x_0$ is a path ending at $\widetilde x_1$. By hypothesis, $H\subseteq g^{-1}Hg$, which leads to the existence of a unique lift $f:\pi_1(\widetilde X,\widetilde x_0)\to\pi_1(\widetilde X,\widetilde x_1)$ such that $p=p\circ f$. This lift turns out to be a surjective covering map itself, and it is a homeomorphism iff $H=g^{-1}Hg$.

I was unsuccessful in showing the injectivity. If $x_1$ and $x_2$ have the same image under $f$, then $x_1$, $x_2$, and $f(x_1)=f(x_2)$ are all in the same fiber. I took $\lambda$ to be a path from $x_1$ to $x_2$. I have been playing around with $\lambda$, $p\lambda$, and $f\lambda$, but got nowhere.

Of course, there could also be a direct algebraic proof. On the other hand, if the statement is not true then someone maybe knows of a counterexample.

share|cite|improve this question

marked as duplicate by user1729, PVAL, Mark Bennet, Jonas Meyer, Hagen von Eitzen Oct 29 '14 at 21:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Some counterexamples are given on MO. – anon Feb 23 '13 at 17:25
up vote 8 down vote accepted

Another example. Let $$ K = \left\{ \frac{a}{2^{n}} : a \in \mathbf{Z}, n \in \mathbf{N} \right\} $$ be the additive subgroup of $\mathbf{Q}$. The map $g : x \mapsto 2 x$ is an automorphism of $K$. Consider the semidirect product $G = K \rtimes \langle g \rangle$. (So that conjugating an element $x$ of $K$ by $g$ in $G$ is the same as taking the value of $g$ on $x$.) Let $H = \left\{ \frac{a}{2} : a \in \mathbf{Z} \right\}$ be a subgroup of $G$. Then $H^{g} = \mathbf{Z} < H$.

PS When I was first exposed to these examples, what struck me is what happens if you look at it from the other end: $\mathbf{Z}^{g^{-1}} = H > \mathbf{Z}$.

share|cite|improve this answer
Thanks for your answer. But I don't quite understand what a group $G$ is. Of what form are the elements? – Stefan Hamcke Feb 23 '13 at 18:07
@StefanH., it's an instance of a semidirect product, an important and useful construction in group theory. Sorry, I am fixing dinner and I do not have time to elaborate, but please take a look at – Andreas Caranti Feb 23 '13 at 18:23
Okay, I didn't know you can take the semidirect product of arbitrary groups (as long as you can assign an automorphism to each element linearly), but here it is quite canonically. I only knew about the internal semidirect product as in the dihedral group. I really like your example! – Stefan Hamcke Feb 23 '13 at 21:46
Thanks @StefanH., the (external) semidirect product is a very handy tool, worth mastering. – Andreas Caranti Feb 24 '13 at 19:05

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