# Left action and group automorphism

I know a group acts on itself by conjugation forming its inner automorphism group. At the same a group acts transitively onto itself by the left action, forming the basis for the Cayley theorem.

I was wondering if a left action can form any automorphism group as well, since I never heard about that (and probably my question doesn't make any sense, sorry in case, I was just trying to follow my intuition).

Thanks

• If by "left action" you mean multiplication on the left, no, since the action maps $e$ to some non-identity element. If you just mean some action on the left, then conjugation $g\cdot x \mapsto gxg^{-1}$ is a left action of $G$ on itself which induces automorphisms. Jul 15 at 16:49
• @ArturoMagidin, I really meant multiplication on the left. So you answered to me currently, tx. Eventually promote it to an answer, tx. Jul 15 at 16:51
• So basically that's not even an homomorphism since doesn't map the identity to itself, right? Jul 15 at 17:17
• Yes.${}{}{}{}{}{}$ Jul 15 at 17:22

To esentialy sum up and promote the comment stream to the question itself to an answer:

Let the group under discussion be denoted by $$G$$.

The left (or right) action by $$a \in G$$ given by left (right) multiplication by $$a$$ is not a homomorphism because

$$a(xy) \ne (ax)(ay) \tag 1$$

in general. Indeed, if

$$a(xy) = (ax)(ay), \tag 2$$

then left multiplication by $$a^{-1}$$ yields

$$xy = xay, \tag 3$$

and performing left multiplication by $$x^{-1}$$ yields

$$y = ay, \tag 4$$

which upon right multiplication by $$y^{-1}$$ gives

$$e = a; \tag 5$$

thus, the left action given by left multiplication by $$a$$ is a homomorphism only in the case that $$a$$ is the identity element of $$G$$.

Of course, the corresponding result holds for right actions.