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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?

Supriyo Halder
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1 Answers1

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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 ($g\leftrightarrow g^{-1}$).

anon
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    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 '17 at 16:13
  • Maybe this is a naive question having not been exposed to transport of structure, but in my mind, your example is just using the symbols of the quaternions while the operation is just modular addition in disguise. Can you clarify exactly the distinction between the cyclic group of eight elements and the quaternions? – Hayden Jul 17 '17 at 16:44
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    @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 '17 at 16:47
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    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 $B'={ -i,-j,-k}$. – Jeppe Stig Nielsen Jul 17 '17 at 22:39
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    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. – Jyrki Lahtonen Jul 18 '17 at 18:22
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    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. – Vandermonde Apr 28 '20 at 06:37