Prove that the map of $A \mapsto BAB^{-1}$ is an automorphism of the group of all Special Matrices $SL(\mathbb{R})$

Question

Let $n \geqslant 1$ be an integer. Prove that for all $B \in GL_n(\mathbb{R})$, the map $A \mapsto BAB^{-1}$ is an automorphism of $SL_n(\mathbb{R})$. Where $S$ is the group of matrices with $\det = 1$.

Here is what I have so far:

Claim: if $A$ belongs to $S$, so does $BAB^{-1}$.

Proof: $\det(CD) = \det(C)\det(D)$ and $\det(C) = 1/\det(C^{-1})$ therefore, $\det(BAB^{-1}) = \det(B)\det(A)\det(B^{-1}) = \det(B) \cdot 1 \cdot \det(B^{-1}) = 1$ therefore $BAB^{-1}$ belongs to $S$

Is this proof enough?

Answer 1

Let $$SL_n(\mathbb{R})=\{ A \in GL_n(\mathbb{R}):|\det(A)|=1 \},$$ which is a group under matrix multiplication. Remember that an automorphism is a bijective endomorphism; that is, a biyective homomorphism $\varphi:G \to G$, where $G=(G,*)$ is a group. Now, if $\phi:SL_n(\mathbb{R}) \to SL_n(\mathbb{R})$ is suh that $\phi(A)=BAB^{-1}$ then $$\phi(AC)=B(AC)B^{-1}=BAI_nCB^{-1}=BA(B^{-1}B)CB^{-1}=(BAB^{-1})(BCB^{-1})=\phi(A)\phi(C),$$ so $\phi$ is an homomorphism.

Now, observe that $$\det(\phi(A))=\det(BAB^{-1})=\det(B)\det(A)\det(B^{-1})=\det(B)\det(A)\frac{1}{\det(B)}=\det(A)=\pm 1,$$ so $\phi$ is surjective ($\phi$ is an epimorphism). Finally, if $\phi(A)=\phi(C)$ then $$BAB^{-1}=BCB^{-1}\Leftrightarrow B^{-1}(BAB^{-1})=B^{-1}(BCB^{-1}) \Leftrightarrow AB^{-1}=AC^{-1} \Leftrightarrow (AB^{-1})B=(CB^{-1})B \Leftrightarrow A=C$$ and hence $\phi$ is injective ($\phi$ is a monomorphism). Therefore $\phi$ is bijective ($\phi$ is an isomorphism). And since $\phi$ goes from $SL_n(\mathbb{R})$ ont the same set we conclude that $\phi$ is an automorphism.

NOTE: Here $I_n$ denotates the identity matrix of order $n$.

Answer 2

Called $\gamma_B$ the application defined by $\gamma_B(A)=BAB^{-1}$ for every $A\in SL_n(\mathbb{R})$ you should note that: $1)$ $\gamma_B$ is well-defined, actually the only thing you proved (which can be sufficient if you can assume the following). $2)\ \gamma_B(I_n)=I_n$ (obvious). $3)\ \gamma_B(AA')=\gamma_B(A)\gamma(A')$ (easy). $4)\ \gamma_B$ are all invertible (pretty simple giving the explict inverse). Now, perhaps you can assume pretty much of these facts as known (the conjugation gives arise to an automorphism) but the exercise is too easy (in my opinion) to jump over these assumptions

Answer 3

It's best to separate out the part that has nothing to do specifically with $\operatorname{SL}_n$ and $\operatorname{GL}_n$.

Lemma. If $N \subseteq G$ is a normal subgroup, then for all $g \in G$, $n \mapsto g n g^{-1}$ is an automorphisms of $N$.

Prove this if you don't think it can be assumed, and cite it or allude to it otherwise. Then it just remains to show $\operatorname{SL}_n$ is normal in $\operatorname{GL}_n$, and that is what you did with your determinant argument.