# Give example of two non-isomorphic groups that have same as sets.

Does there exists two non-isomorphic groups with the following properties:

1.Two groups are same as sets.

2.Two groups have the same identity element.

3.Each element has same inverse in each of the groups.

I have done an exercise on isomorphism chapter of J.Gallian's book where all the criteria are fulfilled but they are isomorphic.But is there a non-isomorphic example?

• What's the number of the exercise that you mention? Jul 17, 2017 at 15:26
• What you can search for it is that take a set and define 2 different binary operations on it such that with each operation, the set becomes a group.
– Our
Jul 17, 2017 at 15:27
• I feel like no such example exists, but I can't see how. Jul 17, 2017 at 15:35

Suppose $G$ and $H$ are any two groups of the same size, which have the same number of elements of order $2$. Suppose we write $G=\{e_{\small G}\}\sqcup X\sqcup A\sqcup A'$ and $H=\{e_{\small H}\}\sqcup Y\sqcup B\sqcup B'$, where $X,Y$ are the elements of order $2$ and $A\sqcup A',B\sqcup B'$ are partitions of the other nontrivial elements of $G,H$ such that no two elements in just one of the sets are mutually inverse.

(So, for instance, if $a\in A$ then $a^{-1}\in A'$.)

Any pair of bijections $X\to Y,A\to B$ extends to a bijection $f:G\to H$ by stipulating the conditions and $f(e_{\small G})=e_{\small H}$ and $f(a^{-1})=f(a)^{-1}$ for all $a\in A$. Then $f$ may be used for transport of structure to make $H$ have two group operations with the same inverses.

The smallest example of when two nonisomorphic groups have the same order and the same number of elements of order $2$ are the cyclic group $\mathbb{Z}_8$ and the quaternion group $Q_8$. In this case, we can biject $\{\pm1,\pm i,\pm j,\pm k\}$ with $\{\bar{0},\bar{1},\cdots,\bar{7}\}$ as follows:

$$\begin{array}{|c|c|} \hline \mathbb{Z}_8 & Q_8 \\ \hline \bar{0} & 1 \\ \hline \bar{1} & i \\ \hline \bar{2} & j \\ \hline \bar{3} & k \\ \hline \bar{4} & -1 \\ \hline \bar{5} & -k \\ \hline \bar{6} & -j \\ \hline \bar{7} & -i \\ \hline \end{array}$$

This allows us to turn $Q_8$ into a cyclic group with respect to a different operation $\bullet$, by stipulating $1$ is still the identity with respect to $\bullet$, the element $i$ is a cyclic generator, its first powers are $i\bullet i=j$ and $i\bullet i\bullet i=k$, and all elements' inverses are given by $-x$ (so that $-x\bullet x=1$).

There are two ideas in this answer. The first is that transport of structure tells us two operations on the same set is equivalent to a bijection between two sets with operations. The second is that the bijection we need must be an isomorphism of pointed $\mathbb{Z}_2$-sets, where any group $G$ is a pointed set because the identity is a distinguished element and is a $\mathbb{Z}_2$-set because $\mathbb{Z}_2$ acts on $G$'s underlying set by inverses ($g\leftrightarrow g^{-1}$).

• As an exercise, consider putting an operation $\bullet$ on $\Bbb Z_3\times\Bbb Z_3$ to make it cyclic in a way that the identity and inverses with respect to $\bullet$ are the same as the usual addition operation.
– anon
Jul 17, 2017 at 16:13
• @HaydenJulius Yes, my example is just using the symbols of the quaternions while the operation is just modular addition in disguise. The distinction between the quaternion group $Q_8$ and the cyclic group $\Bbb Z/8\Bbb Z$ is that they have different underlying sets. (OP wanted two group structures on the same set.)
– anon
Jul 17, 2017 at 16:47
• Nice. It is implicit in your answer that if the two groups do not have the same number of order-2 elements, there is no hope, because of the requirement about "same" inverse. Then your answer shows that this is the only obstacle. For those who do not get it: In your table with $\mathbb{Z}_8$ and $Q_8$, we have: $X=\{\bar{4}\}$ and $Y=\{ -1\}$, and then $A=\{\bar{1},\bar{2},\bar{3}\}$ and $B=\{ i,j,k\}$, and finally $A'=\{\bar{7},\bar{6},\bar{5}\}$ and $B'=\{ -i,-j,-k\}$. Jul 17, 2017 at 22:39
• Sorry about missing your comment re $\Bbb{Z}_3\times\Bbb{Z}_3$. +1 has, of course, been there ever since I read your answer earlier today. Jul 18, 2017 at 18:22
• I came here to say I remember once ending an exchange with something like "$\mathbb{Q}$ is non-abelian because you never specified a group operation so I choose it to be what I want". The moral I was ostensibly supposed to take away was that when most refer to such a thing as "the rationals" they don't mean it just as a set of untethered, meaningless labels (or that I'm too eager to "flout convention", as they say), but I like your answer and it's interesting that technically, one could Frankenstein such a structure onto it. Apr 28, 2020 at 6:37