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Have a doubt in exercise that is repeated in two texts (as stated at the end)-
Suppose there are six elements $1,a,b,c,d,e$ whose laws of combination are given by the table. As per the table, there is identity and inverse; with implied closure under the binary operation. Hence, these elements form a group. Such elements can be shown by two examples:

(i) Permutations: $1, (xyz), (xzy), (yz), (xy), (zx)$;
(ii) Substitutions: $f(x)=x,\frac{1}{1-x},\frac{x-1}{x},\frac 1x,\frac{x}{x-1}, 1-x$

Have table copied from the second source, although the first source has different table. $$ \begin{array}{c|lcr} & \text{f1} & \text{f2} & \text{f3} & \text{f4} & \text{f5} & \text{f6}\\ \hline f1 & f1 & f2 & f3 & f4 & f5 & f6\\ f2 & f2 & f1 & f4 & f3 & f6 & f5\\ f3 & f3 & f6 & f1 & f5 & f4 & f2\\ f4 & f4 & f5 & f2 & f6 & f3 & f1\\ f5 & f5 & f4 & f6 & f2 & f1 & f3\\ f6 & f6 & f3 & f5 & f1 & f2 & f4\\ \end{array} $$



Taking the simpler case of (ii) (also, this example is in both sources), have difficulty in understanding all the entries; but let us consider only the diagonal entries; starting from the left hand corner.
Say, how it is possible that $f1*f1 = f1$, i.e. $x*x = x$. Similarly, how $f2*f2 = f1$, i.e. $\frac 1x*\frac 1x = x$. In fact, none seems to fit the table given.
Instead, what seems to lack is that there should be an additional element for identity function = f0 =1.

Only then can have the concept of inverse, else no function ($f1\cdots f6$) can have any inverse.
Also, even after this modification (of adding, Identity element); the property of closure still does not seem to fit, as the multiplication operation will yield new functions outside the set of elements in the group (of six functions), say $f1*f1 = x^2$, and so on.

Further, have questions:
1. Is the identity element assumed to be present implicitly in the abovemultiplication table.
2. Why the multiplication table of two sources differ?

Both exercises are given as exercise in the book by Harold Hilton on Groups. Also, the second example is repeated as an exercise in the book by Emil Artin, titled 'Algebra with Galois Theory'.

jiten
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    The operation is function composition, not multiplication. – A. Goodier Mar 31 '20 at 08:47
  • @A.Goodier Please elaborate, by taking single example of $f1\ o \ f1 = f1$ – jiten Mar 31 '20 at 08:49
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    So $f_1(f_1(x))=f_1(x)=x$ implying $f_1 \circ f_1=f_1$ while $f_2(f_2(x))=f_2(\frac1x)=\dfrac1{\frac1x}=x$ implying $f_2 \circ f_2=f_1$ – Henry Mar 31 '20 at 08:50
  • @Henry Please put an answer that also covers the first example of permutations. Its multiplication table is given in the first source, although (cannot explain how) the above table should suffice here too. Please note that both are the first example - for groups, (although, labeled as exercise) in both the texts; so solving them helps to proceed in the texts. – jiten Mar 31 '20 at 08:59
  • For a permutation $f_2=(12)$, of course $f_2\circ f_2=f_1$, with $f_1=(1)$. – Dietrich Burde Mar 31 '20 at 09:08
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    If you know how to multiply permutations (given in cycle notation), jiten, then it's a good exercise for you to fill out the table yourself. – Gerry Myerson Mar 31 '20 at 09:08
  • @GerryMyerson Will try, but request how the multiplication tables of the two sources are the same. – jiten Mar 31 '20 at 09:10
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    For each multiplication table, you can pick an element of order $2$, call it $r$, and an element of order $3$, call it $s$, and then use $r^2=1,s^3=1,rs=s^2r$ to fill out the table. So they are both the same table, just with different names for the elements. – Gerry Myerson Mar 31 '20 at 09:13
  • @GerryMyerson Please elaborate it, or better make an answer. Thanks in anticipation. – jiten Mar 31 '20 at 09:16
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    No, jiten, you elaborate it. You work on it until you understand it, and then you post an answer. I have given you everything you need. Now you get to do the work to learn it. – Gerry Myerson Mar 31 '20 at 09:20
  • @GerryMyerson I know that order $=$ # elements in a group, & given $r^2 = s^3 = 1$, I confuse it with degree of group of permutations; as also given here : https://math.stackexchange.com/q/1405460/424260. – jiten Mar 31 '20 at 09:26
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    Order of a group is the number of elements in the group, but order of an element is the smallest nonzero power of that element that gives the identity element. I suggest you take a text and read it from the start and not skip over anything – you must learn these definitions, or you'll get nowhere. – Gerry Myerson Mar 31 '20 at 09:34
  • @GerryMyerson Please make it more explicit how $r^2 = s^3=1$ will help. Also, how is it that it is certain to find $r, s$ in each of the two tables. I mean that does having a group's table means that elements with all orders till some particular value (that, I guess is: group order) can be found. – jiten Mar 31 '20 at 17:17
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    The group table tells you what $pq$ is for all $p.q$ in the group, so it contains all the information there is to be had about the group operation. Please, please, please read through your text – some text – any text – to learn the basic facts about group tables and how to use them to find the orders of elements. I can't give you a course in group theory here, you have to either go through it on your own or if you can't do that get someone – hire someone, if you have to – to go through group theory with you step by step. Start at the beginning. – Gerry Myerson Mar 31 '20 at 22:25
  • Approach of GerryMyerson can be understood (at least partially) by the post at: https://math.stackexchange.com/questions/1008610/an-element-of-a-group-has-the-same-order-as-its-inverse. Also, found a relevant image from the book of Hans Zassenhaus' book on groups; at pg. 9; as shown at: https://i.stack.imgur.com/qCoXx.png The only change needed is to have $d =<1,3>$ , rather than <3, 1>. – jiten Apr 04 '20 at 07:40

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