1. ## Conjugate subgroups

Let G be a group and let g be some element of G. Given that $\displaystyle \tau$ is an isomorphism is such that $\displaystyle \tau(g) = g_1$, show that $\displaystyle S(g)$ and $\displaystyle S(g_1)$ are conjugate subgroups of G. Here, $\displaystyle S(x)$ is the subgroup of the set of automorphisms of G that fix the element x. In other words, $\displaystyle S(x) = \{\sigma\in Aut G \bigr| \ \sigma(x) = x\}$.

I haven't gotten anywhere with this after about 5 hours now. I feel like the key to the problem will be pulling in $\displaystyle \tau$ somehow- intuitively, g and g1 seem to me sort of like inverses of each other across the $\displaystyle \tau$ isomorphism. Since $\displaystyle \tau$ is a member of the group Aut(G), it has an inverse, say $\displaystyle \tau^{-1}$ such that $\displaystyle \tau^{-1}(g_1) = g$ and I feel like $\displaystyle \tau$ somehow will be very involved in the conjugation of $\displaystyle S(g)$. I don't know. I've been staring at a whiteboard basically writing the definition of conjugation over and over and over and over without getting anywhere. I already have proved that $\displaystyle S(g)=\{\sigma\in Aut G \bigr| \ \sigma(g) = g\}$ is a subgroup of Aut(G). I'm stuck :/

2. ## Re: Conjugate subgroups

Originally Posted by cubejunkies
Let G be a group and let g be some element of G. Given that $\displaystyle \tau$ is an isomorphism is such that $\displaystyle \tau(g) = g_1$, show that $\displaystyle S(g)$ and $\displaystyle S(g_1)$ are conjugate subgroups of G. Here, $\displaystyle S(x)$ is the subgroup of the set of automorphisms of G that fix the element x. In other words, $\displaystyle S(x) = \{\sigma\in Aut G \bigr| \ \sigma(x) = x\}$.

I haven't gotten anywhere with this after about 5 hours now. I feel like the key to the problem will be pulling in $\displaystyle \tau$ somehow- intuitively, g and g1 seem to me sort of like inverses of each other across the $\displaystyle \tau$ isomorphism. Since $\displaystyle \tau$ is a member of the group Aut(G), it has an inverse, say $\displaystyle \tau^{-1}$ such that $\displaystyle \tau^{-1}(g_1) = g$ and I feel like $\displaystyle \tau$ somehow will be very involved in the conjugation of $\displaystyle S(g)$. I don't know. I've been staring at a whiteboard basically writing the definition of conjugation over and over and over and over without getting anywhere. I already have proved that $\displaystyle S(g)=\{\sigma\in Aut G \bigr| \ \sigma(g) = g\}$ is a subgroup of Aut(G). I'm stuck :/
If you conjugate an element of $\displaystyle S(g)$ by $\displaystyle \tau^{-1}$, you get an element of $\displaystyle S(g_1)$. Similarly, conjugating an element of $\displaystyle S(g_1)$ by $\displaystyle \tau$ gives an element of $\displaystyle S(g)$. Conjugation by a fixed element is a 1-1 mapping and the result follows.

3. ## Re: Conjugate subgroups

Hi,
You really should be more careful in your posting of a question. It should read something like:

Let $G$ be a group and $g\in G$. Define the stabilizer $S(g)$ of $g$ in $\text{Aut}(G)$ by $S(g)=\{\sigma\in\text{Aut}(G)\,:\,\sigma(g)=g\}$. For any $\tau\in\text{Aut}(G),\,\,S(\tau(g))=\tau S(g)\tau^{-1}$. That is, $S(g)$ and $S(\tau(g))$ are conjugate subgroups of $\text{Aut}(G)$.

The proof is almost trivial: Let $\sigma\in S(\tau(g))$ Then $\tau(g)=\sigma(\tau(g))$ and so $g=(\tau^{-1}\sigma\tau)(g)$ or $\tau^{-1}\sigma\tau\in S(g)$. That is $\sigma=\tau(\tau^{-1}\sigma\tau)\tau^{-1}\in\tau S(g)\tau^{-1}$. Conversely, let $\sigma\in\tau S(g)\tau^{-1}$. Then $\tau^{-1}\sigma\tau\in S(g)$ or $\tau^{-1}\sigma\tau(g)=g$. Thus $\sigma(\tau(g))=\tau(g)$ and $\sigma \in S(\tau(g))$.

The same proof shows that in any permutation group G acting on a set $\Omega$, the stabilizer subgroups of $i$ and $g(i)$ are conjugate for any $i\in\Omega,\,\,g\in G$.

Personal lament. In the "old" days we wrote functions on the right and frequently as exponents. To me, the above proof is much easier with this notation. $i^{gh}=i^g$ if and only if $i^{ghg^{-1}}=i$