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It is shown in section $1.3$ of ug level book on Group Theory by Louis W. Shapiro, a way to build up the possible multiplication tables for groups of order $4$.

I know three properties only, as stated below (also, the reason other posts are unhelpful):
1. In abelian groups, if $ab=c$, then $ba =c$. This implies that their group tables must be symmetrical about the diagonal.
2. If take $ab=a$, then $b=e$; which is wrong.
3. In a group table, it is not possible to have $ab=ac=d$, as it implies $b=c$. Similarly, not possible to have $ba = ca = d$.

The book states two cases based on the choice of $ab = e$, or $ab = c$.

Case I: $ab = e$

$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} & e&a&b&c \\ \textbf{a} &a&&& \\ \textbf{b} &b&&&\\ \textbf{c}&c&&&\end{array}\right]$$

$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} & e&a&b&c \\ \textbf{a} &a&&e& \\ \textbf{b} &b&&&\\ \textbf{c}&c&&&\end{array}\right]$$

$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} &e &a&b&c \\ \textbf{a} &a&c&e&b \\ \textbf{b} &b&&&\\ \textbf{c}&c&&&\end{array}\right]$$

$$Case \ I :\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} &e &a&b&c \\ \textbf{a} &a&c&e&b \\ \textbf{b} &b&e&c&a\\ \textbf{c}&c&b&a&e\end{array}\right]$$

Doubt -1: This is an abelian group, as per the property (#$1$) stated above.
The abelian property of the group is corroborated in the book, as shown in the image below.
But, the diagonal elements in the group table are not identity elements.
Also, no swap of rows / columns can achieve the diagonal elements being identity here.

Case II: $ab = c$

$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} & e &a &b&c \\ \textbf{a} &a&&c& \\ \textbf{b} &b&&&\\ \textbf{c}&c&&&\end{array}\right]$$

$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} & e &a &b&c \\ \textbf{a} &a&&c&b \\ \textbf{b} &b&&&\\ \textbf{c}&c&&&\end{array}\right]$$

$$Case \ II \ (a) :\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} & e &a &b&c \\ \textbf{a} &a&e&c&b \\ \textbf{b} &b&c&e&a\\ \textbf{c}&c&b&a&e\end{array}\right]$$

Doubt -2:
Book states (in table - Case II(a)) only one choice for $ac=b$, but it is not clear why $ac =e$ could not be a choice.
So, select the option $ac=e$ instead to see what changes it brings:
$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} & e &a &b&c \\ \textbf{a} &a&&c&e \\ \textbf{b} &b&&&\\ \textbf{c}&c&&&\end{array}\right]$$ $$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} & e &a &b&c \\ \textbf{a} &a&b&c&e \\ \textbf{b} &b&&&\\ \textbf{c}&c&&&\end{array}\right]$$

$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} & e &a &b&c \\ \textbf{a} &a&b&c&e \\ \textbf{b} &b&c&&\\ \textbf{c}&c&e&&\end{array}\right]$$

$$Case \ II \ (b) :\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} & e &a &b&c \\ \textbf{a} &a&b&c&e \\ \textbf{b} &b&c&e&a\\ \textbf{c}&c&e&a&b\end{array}\right]$$

Doubt -3:
Case II (c) : It is stated in the book, for Case II, that had we chosen $bc=e$, then would have reverted back to case I.
I want to ask, how does it become case I.

$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} & e &a &b&c \\ \textbf{a} &a&e&c&b \\ \textbf{b} &b&&&e\\ \textbf{c}&c&&&\end{array}\right]$$

As not detailed further in the book, tried to fill the rest of the table as shown below:
No choice apart from $ba=c$, also no choice apart from $bb=a$.

$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} & e &a &b&c \\ \textbf{a} &a&e&c&b \\ \textbf{b} &b&c&a&e\\ \textbf{c}&c&&&\end{array}\right]$$

Similarly, the last row has restrictions as no choice apart from $ca=b$; $cb=e$.

$$Case \ II \ (c) :\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{e} & e &a &b&c \\ \textbf{a} &a&e&c&b \\ \textbf{b} &b&c&a&e\\ \textbf{c}&c&b&e&a\end{array}\right]$$

But, it is unclear how the substitution has made the table like that of Case I.

enter image description here

jiten
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    I cannot see what most of your doubts are, but you do say " that had we chosen $bc=e$, then would have reverted back to case I". This is because you have the product of two distinct non-identity elements being the identity. Concretely, if you relabel $a$, $b$, $c$ as $c$, $a$, $b$ you are back in case I. – Angina Seng Apr 07 '20 at 03:23
  • @AnginaSeng Failed to understand your comment that states to convert table for Case I to that to Case II(c), by renaming. Request to elaborate, as for me both are unique, and do not know how to implement your comment. – jiten Apr 07 '20 at 03:33

2 Answers2

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Doubt 1: having all identity elements on the diagonal is not what it means to be abelian - a group is abelian if its multiplication table is symmetric about the diagonal, since that says $xy = yx$ for all $x$ and $y$.

Doubt 2: if $ac=e$ then rename so that $c$ becomes $b$, and then you return to the first case.

Doubt 3: this is the same as Doubt 2, if $bc = e$ then we could change the names at the beginning to instead have $ab=e$.

The point for both Doubts 2 and 3 is that the assignment of the labels $a,b,c$ to the elements of the group was arbitrary, and we can always reshuffle the names. For instance, try working out $bc= e$ by taking the proof of Case 1 and replacing every "b" by "a", every "c" by "b" and every "a" by "c".

  • Request showing how the renaming of labels will help - i.e. for Doubts -2,3. I would be highly thankful for this help. – jiten Apr 07 '20 at 03:35
  • As I said, you should literally take the proof of case 1 and make the replacement. The names of variables do not affect the results of a proof. I am not sure what I can say that will better convince you. –  Apr 07 '20 at 19:51
  • Thanks. Also, is there an algorithmic approach to find out which of the two rows ( & the corresponding columns) to swap (as the first step, before renaming the variables in the second step) for showing equivalence between two group tables. – jiten Apr 08 '20 at 03:40
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Doubt $1$:

For an Abelian group, we need $xy=yx$, i.e. when $x=y$ (i.e. along the diagonal), we have $x^2=x^2$ but $x^2$ need not be equal to the identity. We do not need all diagonal entries to be identity or this will further impose the condition that every element is its own inverse. This is not a requirement for an Abelian group.

Doubt $2$ and $3$:

View case $(II)(a)$ as the case where every element is its own inverse and case $(I)$ as the case where exactly $2$ elements have itself as its inverse.

Replace the element besides $e$ with itself as its own inverse as the 'sun'. Then for element that are not equal to the identity and the 'sun', we call it the 'moon' and the 'earth'.

Then table $I$, with $c$ as the 'sun':

$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{moon} & \textbf{earth} & \textbf{sun}\\ \hline \textbf{e} &e &moon&earth&sun \\ \textbf{moon} &moon&sun&e&earth \\ \textbf{earth} &earth&e&sun&moon\\ \textbf{sun}&sun&earth&moon&e\end{array}\right]$$

Table $(II)(b)$ with $b$ as the 'sun':

$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{moon} & \textbf{sun} & \textbf{earth}\\ \hline \textbf{e} & e &moon &sun& earth\\ \textbf{moon} &moon&sun&earth&e \\ \textbf{sun} &sun&earth&e&moon\\ \textbf{earth}&earth&e&moon&sun\end{array}\right]\tag{1}$$

As an exercise, fill in the table according to $(1)$

$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{moon} & \textbf{earth} & \textbf{sun}\\ \hline \textbf{e} & & && \\ \textbf{moon} &&&& \\ \textbf{earth} &&&&\\ \textbf{sun}&&&&\end{array}\right]\tag{2}$$

You should be able to see that the two tables are equivalent.

Table $(II)(c)$ has $a$ as the 'sun'.

$$\left[\begin{array}{c|cccc}* & \textbf{e} & \textbf{sun} & \textbf{earth} & \textbf{moon}\\ \hline \textbf{e} & e &sun &earth&moon\\ \textbf{sun} &sun&e&moon&earth \\ \textbf{earth} &earth&moon&sun&e\\ \textbf{moon}&moon&earth&e&sun\end{array}\right]$$

Again, fill in the table and compare with the first table.

Siong Thye Goh
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  • I hope that there must be a better way to see such equivalent tables, as it is not feasible for next bigger tables even (i.e. $5*5$). Also, there are two cases that we need to find equivalence :(i) betn case II(b) and case II(a); (ii) betn case II(c) and case I. I am confused if you have instead taken (i) as showing equivalence between case II(a) and Case I. Also, am still not clear about how the swapping elements are chosen, & request more detail. Have successfully done (ii), but want to edit answer when am clear about how swap elements were chosen in table. Can be in old chat room instead. – jiten Apr 07 '20 at 09:51
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    The last case is left as an exercise for you. Given a positive integer  $n$ , it is not a simple matter to determine how many isomorphism types of groups of order  $n$ there are. If  $n$ is the square of a prime, then there are exactly two possible isomorphism types of groups of order  $n$ , both of which are Abelian. – Siong Thye Goh Apr 07 '20 at 10:09
  • The last case, I hope, is (ii) i.e. showing equivalence betwn. case II(c) and Case I tables. It is done successfully by me ( but will edit answer to show it once am clear). I am stuck in showing the equivalence betwn. case II(a) and case II(b) tables. Thanks for detailing the concept known for showing equivalence between the two tables - i.e. isomorphism types. – jiten Apr 07 '20 at 10:15
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    $II(a)$ and $II(b)$ are not equivalent. – Siong Thye Goh Apr 07 '20 at 10:28
  • I have also solved the case (i) of finding equivalence betn. tables of Case II(b) and Case I. But, it is still not clear to me, how you found that for (i) need swap $b$ and $c$, while for (ii) need swap $a$ and $c$. Will edit, once am clear about that. – jiten Apr 07 '20 at 11:20
  • at least align the sun, the element of which we have identify that it is different from the rest. – Siong Thye Goh Apr 07 '20 at 11:30
  • Please elaborate what is meant by aligning the last row/column, as it is different from the others. In what respect, the difference should be judged? Can we be in chatroom instead? – jiten Apr 07 '20 at 11:33
  • Good question on table filling with missing entries at : https://math.stackexchange.com/q/176234/424260. Please read my comments to that. – jiten Apr 08 '20 at 04:10
  • Please be in chatroom. – jiten Apr 12 '20 at 01:54
  • Please help me with my post at: https://math.stackexchange.com/q/3632815/424260. Am really stuck in it for many days. – jiten Apr 21 '20 at 00:59
  • Please help with my last comment, particularly why two different approaches (with their different answers) are working. – jiten Apr 21 '20 at 01:48
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    I am responding to multiple emails and will get back to you. – Siong Thye Goh Apr 21 '20 at 01:54