# Automorphism and conjugation

Let $G$ be a group and fix $h\in G$. Show that $f:G\to G$ given by $f(g)=h^{-1}gh$ is an automorphism.

Do I need to show that $f$ is bijective first? And then relate it to being an automorphism? A function to be automorphism must be isomorphism, so bijective.

• It shouldn't be hard to show that it's a bijection, since you're looking at cosets of the entire group. After that, you just need to show it is a homomorphism of groups. – TokenToucan Dec 1 '15 at 22:46
• I've edited your post. When quoting text, you can use blockquotes by putting > in front of the line. There are many sources around to help you format mathematical expressions and equations using $\LaTeX$ markup. Also, this has nothing to do with (automorphic-forms), for future reference you can use the tag descriptions to help determine when tags are relevant to your question or not - I replaced it with (group-theory). – whacka Dec 1 '15 at 22:47
• Oh thank you very much for the support! I really appreciate it :) – andy Dec 1 '15 at 22:51
• Either "bijective" or "homomorphism" first is OK, they both need shown, ultimately. For "bijective", note that there is an inverse function; a way to turn $h^{-1}gh$ back into $g$. – pjs36 Dec 1 '15 at 22:56
• Notice that the map $h \mapsto f_h$ isn't a homomorphism in general : math.stackexchange.com/questions/1966720/… – Watson Nov 28 '16 at 20:59

## 2 Answers

First check it's a homomorphism, i.e. $f_h(ab)=f_h(a)f_h(b)$. In fact since $f_h$ has the same source and target it is an "endomorphism." Then you just need to check bijectivity.

Fact: if $f:A\to B$ has an inverse function $g:B\to A$ (i.e. such that $f\circ g=\mathrm{id}_B$ and $g\circ f=\mathrm{id}_A$) then $f$ and $g$ are bijections (one-to-one and onto).

What do you think the inverse function is? If $f_h(x)=y$, can you solve for $x$ in terms of $y$? Once you've guessed the inverse function (or solved for it) verify it is.

For it to be an automorphism it must be a endomorphism and an isomorphism, clearly it's endomorphism because it sends to itself and $$f_h(ab)=habh^{-1}=hah^{-1}hbh^{-1}=f_h(a)f_h(b)$$

To check Isomorphism we check if it's an epimorphism (surjective) and a monomorphism (injective)

Epimorphism is fairly trivial, if $c$ is given then we have that for $a=h^{-1}ch$ satesfies $f_h(a)=c$ so it's an epimorphism.

For monomorphism we can just check that $\ker f_h = e$ which is also easy, when is $f_h(a)=hah^{-1}=e$? We can't have $a=h$ or $a=h^{-1}$ because then we end with $h$ or $h^{-1}$ as a result which are not $e$, but if $a\neq e$ then we still get a non-identity element so we must have $a=e$ and ergo the kernel is trivial.

Therefore it's a monomorphism, which gives it being an isomorphism which gives it being an automorphism.