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EDIT: G is finite.

Let $G$ be a group and $g\in G$. I'm trying to show that the function $\phi_g :G\rightarrow G$ defined by $\phi_g (x)=gxg^{-1}$ for every $x\in G$ is a bijection.

I've shown that $\phi_g$ is injective. Does injectivity along with equal cardinality imply surjectivity and hence a bijection?

Loobear23
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  • https://math.stackexchange.com/questions/1359905/does-same-cardinality-imply-a-bijection – rims Jan 09 '20 at 15:05
  • Are you assuming that $G$ is finite? If so, injective does imply surjective. If not, then it doesn't. – lulu Jan 09 '20 at 15:05
  • Still, a direct proof...given $y\in G$ can you find an $x$ such that $\phi_g(x)=y$? – lulu Jan 09 '20 at 15:07
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    Really not difficult. We want $gxg^{-1}=y$. How can we solve that for $x$? – lulu Jan 09 '20 at 15:10
  • With your edit, the answer is yes. The image is a subset of $G$ with the same cardinality as $G$, so it must equal $G$. – almagest Jan 09 '20 at 15:12

2 Answers2

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No, you do not have such an implication. For instance, in the group $\Bbb Z$ with addition, the homomorphism given by multiplication by $2$ is injective and equal cardinality, but it is not a bijection.

In this case, I would rather try to find an explicit inverse (i.e. guess what the inverse of $\phi_g$ ought to be, then show that that is indeed an inverse).

Responding to the edit: If $G$ is finite, then any injection (homomorphism or not) is indeed a bijection.

Arthur
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The right answer, I think, has been given by lulu above. We do not need the edit that $G$ is finite to show that the map $\phi_g$ is bijective. It is injective and certainly surjective, since $gxg^{-1}=y$ implies that $x=g^{-1}yg$. Of course, if one does not see this, one can ask if injectivity is enough. This is not the case in general as Arthur has pointed out.

More generally, $\phi_g$ is an inner automorphism of $G$ for all $g\in G$.

Dietrich Burde
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