As a prequel to Problem 56 on this problem set
- Let $H$ and $K$ be subgroups of a finite group $G$ and assume $H$ is isomorphic to $K$. Prove that there exists a group $\bar G$ containing $G$ as a subgroup, such that $H$ and $K$ are conjugate in $\bar G$.
I've decided to start with a simplified version:
Given a group $G$ and $\phi\in\text{Aut}(G)$, find a group $\bar G\geq G$ such that there exists $\psi\in\text{Int}(\bar G)$ such that $\psi|_G=\phi$.
In reality Problem 56 doesn't imply my statement so I'm not even sure if it's true! but if it is then I find it cool.
My attempt goes as follows.
Construct $\bar G=\bigcup_{k\in\mathbb{Z}}Gx^k$ for some external $x$. Define, for all $g\in G$, $$xg=\phi(g)x$$ or equivalently $$x^kg=\phi(g)x^k\quad\forall\,k\in\mathbb{Z}.$$
I'd say this operation, together with the standard product from $G$, endows $\bar G$ with a group structure. Moreover for all $g\in G$ we have that $$^xg=\phi(g)$$ so that $\psi_x|_G=\phi$.
To prove that my operation is well defined is what I'm not so sure of how to do. Is it fine if I take $y\in \bar G$ and argue that it's uniquely expressed as $y=gx^k,g\in G,k\in\mathbb{Z}$, and that my operation in $\bar G$ (which I'll denote $\cdot$) really means $$(gx^k)\cdot (hx^\ell)=g\phi^k(h)x^{k+\ell}$$ and now prove associativity, existence of neutral, etc?
We've never actually constructed a group from another in class. Also, is there a standard way to define $\bar G$, such as $$\{gx^k:g\in G,k\in \mathbb{Z},xg=\phi(g)x\}$$ or something along those lines?