Suppose you have two isomorphic groups. Does the binary operation defined on each group need to be the same operation?


closed as off-topic by José Carlos Santos, Gibbs, Lord Shark the Unknown, Lee David Chung Lin, Joonas Ilmavirta Jan 31 at 7:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Gibbs, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ If you mean, does it have to be the same operation in both groups: no. $\endgroup$ – David Jan 30 at 23:48
  • $\begingroup$ Short answer" "no". Possibly helpful: math.stackexchange.com/questions/2039702/… $\endgroup$ – Ethan Bolker Jan 31 at 2:35
  • $\begingroup$ Not anymore than two isomorphic groups are equal $\endgroup$ – BallpointBen Jan 31 at 5:50
  • $\begingroup$ I guess if you think of it strictly in terms of functions then the two operations are defined on different domains so they wouldn't be the same. I was just thinking that if one group had addition (like defined with integers), then would an isomorphic group need to have the same rule. The inspiration for this post came from an assignment question I was doing, showing that any non-cyclic group of order 4 must be isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. $\endgroup$ – NickBlotted Feb 1 at 0:19

Not necessarily, no.

The two binary operations behave the same on the underlying sets with respect to the axioms of group theory, yeah, and that's by the fact that the two groups are isomorphic as groups, but it might so happen that, say, one binary operation is the restriction of some bigger relation whereas the other is not.

For example, $(\Bbb Z_3, +_3)\cong H$, where $H=(X, \circ')$, for $X=\{\operatorname{id}, \sigma^2, \sigma^4\}$, is the subgroup of the group $(\{\sigma^i\mid i\in\overline{0,5}\}, \circ)$ of the rotational symmetries of this shape:

Symmetric tiling under Z_6,

where $\sigma$ is a clockwise rotation of $\pi/3$ radians about the centre, so that $\circ'=\circ\rvert_{X\times X}$ is distinct from addition $+_3$ modulo $3$.

Note that here we have simply declared that $\circ'$ is some restriction but $+_3$ is not (although it can be seen as such); also, $\circ'$ is composition of functions, whereas $+_3$ is an arithmetical operation.

The underlying sets can be completely different too (as illustrated above). A bijection between sets is not necessarily an equality; think: permutations.

Image source.


Consider this:

$\log: (\mathbb R^+,\times) \to (\mathbb R,+)$ is a group isomorphism.


$G_1 = \{0\}$ and $G_2 = \{1\}$ are groups (each with multiplication as the binary operation).

They are isomorphic, as you can show.

The binary operation of $G_1$ has as its domain $\{(0,0)\}$, while the binary operation of $G_2$ has as its domain $\{(1,1)\}$, so their binary operations are not equal.


To add yet another example to those above, note that if $\mathcal{P}_n$ denotes the group of $n$-by-$n$ permutation matrices (binary operation is matrix multiplication) and $S_n$ denotes the symmetric group (binary operation is function-composition), then these groups are isomorphic via the isomorphism $\sigma \longmapsto P_\sigma$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.