# Does the operation of a group matter when talking about isomorphism?

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?

• 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 Nov 5, 2015 at 11:16

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.
• "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 $ℝ^1$. Nov 5, 2015 at 16:39
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.