If $(G,\cdot)$ and $(G,+)$ are groups, are they always isomorphic?

Question

It is possible for a group to have alternative operation. For example, consider the following operations on $\mathbb{Z}^2$: $(x_1,y_1)\cdot(x_2,y_2)=(x_1+x_2,y_1+y_2+x_1x_2)$ and $(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)$. Both $(\mathbb{Z}^2,\cdot)$ and $(\mathbb{Z}^2,+)$ are groups, and I know that they are isomorphic. However, is it true that in general, a group is isomorphic to itself under an alternative operation? ie, if $(G,\cdot)$ and $(G,+)$ are groups, are they always isomorphic? I can't think of an counterexample.

Answer 1

No.

Think of $G=\{e,a,b,c\}$ as a set.

Define

$$\begin{array}{c|cccc} \star & e & a & b & c\\ \hline e & e & a & b & c\\ a & a & b & c & e\\ b & b & c & e & a\\ c & c & e & a & b \end{array}.$$

Then $(G,\star)\cong \Bbb Z_4$.

Define

$$\begin{array}{c|cccc} \ast & e & a & b & c\\ \hline e & e & a & b & c\\ a & a & e & c & b \\ b & b & c & e & a\\ c & c & b & a & e \end{array}.$$

Then $(G,\ast)\cong \Bbb Z_2^2$.

Answer 2

To clarify your question, I think you mean if a set $S$ admits two group operations must they be isomorphic? That is clearer than considering a "group to have an alternate operation". In that case there are simple counterexamples, e.g. a set with 4 elements admits two distinct group structures: one is isomorphic to $\mathbb{Z}/4$ and the other to $\mathbb{Z}/2 \oplus \mathbb{Z}/2$. While writing this the above equivalent answer was posted.