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Does a bijective map between two groups always produce an isomorphism?

I am trying to find a bijective map between two groups which does not preserve the group operations.

I have found a bijection $f(g)=g^3$, where $f:\mathbb{R}\rightarrow\mathbb{R}$, defined by $f(g+h)=(g+h)^3$.

But $f(g + h) \neq $ $ f(g)+f(h) \\=g^3+h^3$

Martin Sleziak
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Astrolink
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    $f:\Bbb Z_2\to \Bbb Z_2: 0\mapsto 1, 1\mapsto 0$. – walcher Sep 12 '13 at 08:53
  • by its definition, an isomorphism is a "homomorphism" which is bijective.. –  Sep 12 '13 at 09:06
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    @PraphullaKoushik Actually, not by its definition. That's a theorem. A theorem which is untrue in other concrete categories :) – Alex Youcis Sep 12 '13 at 09:09
  • I do not know about categories... at least in Group theory, I learned that an isomorphism is a homomorphism which is a bijection.... –  Sep 12 '13 at 09:11
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    @PraphullaKoushik Technically it's defined to be a homomorphism which is invertible, with inverse also a homomorphism. It's a theorem that if f is bijective its inverse is automatically a homomorhpism, but this isn't the definition (or at least not the right definition). – Alex Youcis Sep 12 '13 at 09:21
  • @AlexYoucis : yes, yes.. It seem to be true.. :) :) –  Sep 12 '13 at 09:29
  • @Astrolink The map you give is not a bijection. What is mapped to $2\in\mathbb{Z}$? – user1729 Sep 12 '13 at 09:33
  • @Astrolink Your edit now means that you have answered your own question! You have found a bijection between groups which is not a homomorphism, although it does preserve the identity (and so it is closer to Tobias' answer than to azimut's and walcher's ones, where they shift the identity). – user1729 Sep 12 '13 at 09:38
  • @Alex, will you please provide a reference to a bijective morphism which fails to meet this criterion (it can certainly be textbook; I obviously am new to this)? – Jonathan Y. Sep 12 '13 at 09:39
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    @JonathanY. There are lots of examples. The map $\mathbb{A}^1\to V(y^2-x^3)$ defined by $t\mapsto (t^2,t^3)$ is a bijective morphism of varieties, but not an isomorphism (it certainly can't be by smoothness issues), the map $X\to Y:x\mapsto x$ where $X=\mathbb{R}$ with the discrete topology and $Y=\mathbb{R}$ with the usual topology is continuous and bijective, but not an isomorphism, etc. – Alex Youcis Sep 12 '13 at 09:42

4 Answers4

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For a group $G$ with $\lvert G\rvert > 1$, we can find a bijection from $G \to G$ that's not a homomorphism.

Since the order of $G$ is at least $2$, there is a $g\in G$ with $g\neq 1_G$.

Now the map $f_g : G \to G$ $$ f_g : x \mapsto \begin{cases} g & \text{if } x = 1_G,\\ 1_G & \text{if }x = g,\\ x & \text{otherwise} \end{cases} $$ is a bijection, but not a homomorphism (since $f_g(1_G) \neq 1_G$).

user1729
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azimut
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Interestingly, there are a few groups where any bijective map from the group to itself which preserves the identity element are automorphisms (ie, isomorphisms of the group with itself).

These groups are precisely $\{1\}$, $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/3\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$.

It is easy to check that these groups satisfy the above property, and that these are the only ones follows from my answer at Is a Bijection From a Group to Itself Automatically an Isomorphism If It Maps the Identity to Itself?

Community
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Tobias Kildetoft
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This is by means of azimut's answer. Black is group G. Orange is x.

enter image description here

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A maybe more "natural" example is the following map: if $G$ is not trivial, pick $g\in G,g\neq 1_G$.

Then the map $$\ell_g:x\in G\mapsto gx\in G$$ is bijective (with inverse $\ell_{g^{-1}}$), but not a group homomorphism, since $g=f(1_G)=f(1_G ^2)\neq f(1_G)^2=g^2$ (otherwise, by simplification by $g$, we would have $g=1_G$)

GreginGre
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