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).


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    $\begingroup$ 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. $\endgroup$ Jul 15 at 16:49
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    $\begingroup$ @ArturoMagidin, I really meant multiplication on the left. So you answered to me currently, tx. Eventually promote it to an answer, tx. $\endgroup$ Jul 15 at 16:51
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    $\begingroup$ So basically that's not even an homomorphism since doesn't map the identity to itself, right? $\endgroup$ Jul 15 at 17:17
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    $\begingroup$ Yes.${}{}{}{}{}{}$ $\endgroup$ Jul 15 at 17:22

1 Answer 1


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.


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