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I am starting to study Group Theory and I'm having a problem in defining what is a isomorphism in relation to an homomorphism.

Consider a mapping $f : G_1 \to G_2$ between two groups.
Considering I know this mapping represents an homomorphism AND a bijection.

Question 1:
Is it true that i dont know yet if it represents an isomorphism? Do I have to check if it's inverse is also a bijection?

I was told the groups $(\operatorname{GL}_n(\mathbb{R}) , \times )$ and $(\mathbb{R}^*, \times )$ are not isomorphic, that is, there is not any possible isomorphism between them because one is abelian whereas the other isn't.
At the same time I was told that there can be bijective mappings between these two.
Question 2 :
Does anyone know any example of bijective homomorphisms between those two?

Thanks

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    $\begingroup$ Any bijective map which is a homomorphism is automatically an isomorphism. The inverse is clearly bijective (basic set theory exercise), and that it is a homomorphism it not too hard to check (a good exercise). $\endgroup$ Apr 29, 2013 at 17:15
  • $\begingroup$ I rolled back your edit because you were attempting to change the question completely. If you want to ask another question, do a fresh one. $\endgroup$
    – rschwieb
    Apr 29, 2013 at 17:39
  • $\begingroup$ That`s fair, i didnt change it back because i had lost all my text. $\endgroup$
    – nerdy
    Apr 29, 2013 at 17:42
  • $\begingroup$ For Q1, there could be a bijective function between the two, but there cannot be a bijective homomorphism. The bijective function will not preserve multiplication. $\endgroup$ Apr 29, 2013 at 20:47

2 Answers 2

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  1. Usually, isomorphisms for groups, rings, vector spaces, modules etc are defined to be bijective homomorphisms. However, if your definition of isomorphism $f$ is that there is another homomorphism $g$ such that $fg$ and $gf$ are identity maps, then Tobias Kildetoft's comment on your post provides a full explanation for that. That is, you can prove that the inverse mapping of a bijective homomorphism is also a homomorphism.

  2. It's very easy to disprove that $GL(n,\Bbb R)$ is not isomorphic to $(\Bbb R^\ast,\times)$: you just have to think of a group behavior possible in one and not in the other! The one that came to mind for me is that there are exactly two things in $\Bbb R^\ast$ which square to 1, namely $1$ and $-1$. But in $GL(2,\Bbb R)$, for example, it is very easy to find more than two things which square to 1:

$$ \begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}-1&0\\0&1\end{bmatrix},\begin{bmatrix}-1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\1&0\end{bmatrix}\dots $$

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  • $\begingroup$ Sorry, i had solved my doubt i edited my question before i saw you had answered it. $\endgroup$
    – nerdy
    Apr 29, 2013 at 17:33
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this is some kind of hints:

1) the inverse of a bijection is a bijection (just apply the definition, you can change the role of $f$ with $f^{-1}$, otherwise you can use the fact that bijection is an equivalence relation)

2) the bijection is a function giving an exact pairing of the elements of two sets. We do not assume NOTHING on "is this function preserving the structure and properties of the 2 sets? a bijective homomorphism in fact PRESERVE qualities of a set, for example, if a group called "A" is isomorphic to $\mathbb{Z}_p$ with $p$ prime, we know that A is a cyclic group of order $p$ (and many other things). Just because exist a bijective homomorphism between the 2 groups.

hope this solves some of your doubts

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  • $\begingroup$ Sorry, i had solved my doubt i edited my question before i saw you had answered it. $\endgroup$
    – nerdy
    Apr 29, 2013 at 17:33

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