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When we say two groups are isomorphic to each other, do we have to specify the operations on each group? Does this matter? For example, here is a problem from the book 'Algebra and Geometry' by A.F.Beardon:

Let $G$ be the group of real $2\times 2$ matrices of the form $$M(a)= \begin{pmatrix} a & a \\ a & a \\ \end{pmatrix} $$ where $a \neq 0$.

Show that $G$, under the usual multiplication of matrices, is isomorphic to $\mathbb R^*$, the group of non-zero real numbers.

I solved this problem by defining an isomorphism $$\theta(M(a))=2a$$ which does not involve any specification of the operation on group $\mathbb R^*$.

So my question is : When we talk about the isomorphism between two groups, is it important to specify also the operations defining the two groups?

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    $\begingroup$ Yes, of course. Else you are only talking about bijections. For instance, $\mathbb{Z}/6\mathbb{Z}$ and $S_3$ are non-isomorphic as groups, but there is a bijection between them $\endgroup$
    – Prahlad Vaidyanathan
    Commented Nov 5, 2015 at 11:16

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Yes, it doesn't make sense to talk about a map of sets being an isomorphism of groups unless you have a group structure specified on both the source and the target.

With that being said, often there's only one "reasonable" group structure on a set, and if the group structure is not specified, we always mean that one. For example, the only reasonable group structure on $\mathbb{R}^\times$ is multiplication, and your book would have written a sentence explaining what it meant if it meant otherwise.

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  • $\begingroup$ "For example, the only reasonable group structure on $ℝ^x$ is multiplication" - strange, I would think it's addition since per-component multiplication isn't invariant to rotations. For $ℝ^3$ we also have the cross product, but that's a special case (and so is ordinary multiplicationi n $ℝ^1$. $\endgroup$
    – John Dvorak
    Commented Nov 5, 2015 at 16:39
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    $\begingroup$ Oh. The OP meant the group of non-zero real numbers. Then I agree... $\endgroup$
    – John Dvorak
    Commented Nov 5, 2015 at 16:41
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You assert that $\theta$ is an isomorphism. But, how did you check that this assertion is true? What you had to do, in order to verify the definition of isomorphism, is to verify the equation $\theta(M(a) \cdot M(b)) = 2a \cdot 2b$. That "$\cdot$" character on the left hand side is the group operation in $G$, and the "$\cdot$" character on the right hand side is the group operation in $\mathbb{R}^*$. So it does involve specifying the group structure on the source and the target.

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