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This is a question from Dummit and Foote. I am still a novice at algebra so any feedback on my work would be appreciated!

Let $G$ be a group and let $G$ act on itself by left conjugation, so each $g \in G$ maps $G$ to $G$ by

\begin{align*} x\mapsto gxg^{-1} \end{align*} For fixed $g \in G$, prove that conjugation by $g$ is an isomorphism from $G$ onto itself (i.e. an automorphism of $G$). Deduce that $x$ and $gxg^{-1}$ have that same order for all $x \in G$ and that for any subset $A$ of $G$, $\vert A \vert=\vert gAg^{-1}\vert$ (here $gAg^{-1}= \{gag^{-1}\vert a \in A\}$.)

Take $g \in G$. We show that $\sigma_{g}:G \rightarrow G$ defined by $\sigma_{g}(x) = gxg^{-1}$ is a homomorphism and is bijective. First, we show that $\sigma_g$ is homomorphism. Take $x,y \in G$. \begin{align*} \sigma_g(xy) &= g(xy)g^{-1} \\ &= gxg^{-1}gyg^{-1} \\ &= \sigma_g(x) \sigma_g(y) \end{align*} Hence, $\sigma_g$ is homomorphism. Second, we show that injectivity by showing $ker(\sigma_g) = \{1\}$. Suppose, $gxg^{-1}=1$. Then, \begin{align*} xg^{-1} &= g^{-1} \\ x &= g^{-1}g = 1 \end{align*} Hence, $ker(\sigma_g) = \{1\}$. Now, we show $\sigma_g$ is surjective. Take $g \in G$. Then, for $x = g^{-1}yg \in G$ (since $G$ is a group, $x \in G$), $g(g^{-1}yg)g^{-1} = y$. Therefore, $\sigma_g$ is bijective and it follows that it is a isomorphism. Since $\sigma_g$ is isomorphism, it follows that for every $x \in G$, $|x| = |\sigma_g(x)|=|gxg^{-1}|$. Furthermore, since $\sigma_g$ is bijective on $G$, $\sigma_g$ is bijective on $A \subset G$. In particular, $\sigma_g$ is injective when restricted to $A$. Hence, $|A|= |im(\sigma_{g|A})|=|gAg^{-1}|$.

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  • $\begingroup$ It is ok for me. $\endgroup$ – Éric Guirbal Oct 18 '15 at 21:00
  • $\begingroup$ Only strange line is $|x|=|\sigma_g (x)|$, since you are talking about cardinality (I assume) but $x$ is not a set. $\endgroup$ – Elliot G Oct 18 '15 at 21:14
  • $\begingroup$ @ElliotG For an element $x$ of a group, $|x|$ is the notation used by D&F for the order of $x$. $\endgroup$ – Ovi Sep 27 '17 at 23:16
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All is fine. An alternative way to establish bijectivity might be the observation that $\sigma_g\circ\sigma_h=\sigma _{gh}$ (a useful fact on its own!) and therefore $\sigma_{g^{-1}}\circ \sigma_{g}=\sigma_{g}\circ \sigma_{g^{-1}}=\operatorname{id}_G$. - And a map with left and right inverse map is bijective. Then again, this does not reall ydiffer from what you wrote, does it?

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