University Maths. Intro to Algebra

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I literally got no clue how to answer Q2 and Q3. Also if someone can verify Q1

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i just dont know how to answer q2 and 3

would rlly appreciate the help

[deep forge]

**Explanation of the notation that Q2 introduces: **
given the set {1, 2, 3}, we denote by (abc) the bijection that sends 1 to a, 2 to b, 3 to c,
i.e. (abc) is the application f such that f(1) = a, f(2) = b, f(3) = c,

(where a, b and c represents three distinct elements from {1, 2, 3}) For example, we denote by (321) the application f such that f(1) = 3, f(2)=2, f(3)=1

Composition of two bijections
now if I have two bijections f and g, I can compose them to obtain a new bijection f o g

and the image of an element x through f o g is: (f o g)(x) = f( g(x) )

so to compute the image of 1 by f o g, first compute the image of 1 by g, let's say the result is called d, then compute the image of d by f.
When you've computed the images of 1, 2, and 3 thru f o g, you can now write f o g in the notation Q2 uses, i.e. three numbers in a bracket

Composition table
You now treat this "composition" operation like a multiplication (between two bijections f and g), and you fill each entry of the table by giving the result you obtain by composing the bijection corresponding to the line with the one corresponding to the column

For example the entry at line 3, col 2 is filled.
HOW?

As I mentionned before, you compose the application that corresponds to the line : (213) (let's call it f), with the one that corresponds to the column : (132) (let's call it g)
let's call the resulting bijection: fog = (213)o(132)
We want to write it in this form: (. . .)
For this we need to compute the image of 1, 2, and 3:

**(fog)(1) = f( g(1) ) **
and g(1) is 1 (we know this by looking at g : (132)

so (fog)(1) = f(1), and f(1) is 2 (we know this by looking at f:(213)

so we have (fog)(1) = 2

using the same method, we compute (fog)(2):
(fog)(2) = f(g(2)) = f(3) = 3

and (fog)(3):
(fog)(3) = f(g(3)) = f(2) = 1

we then write (fog) in the desired format :
fog = (fog(1) , fog(2) , fog(3))
= (231)
and fill the table

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thanks so much

im gonna do Q2 now u mind checking if im right or wrong after

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I got this

I need help on Q3 though

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@deep forge

This is wat I got

Could u check if it’s correct for q1,2,3

That’s Q3

This is q2

[deep forge]

sorry for late reply. Q2 is true for as much answers as I looked

[deep forge]

your answers in Q3 are true. A few notes on reasoning:
do not forget "for all" in your definition of an Abelian group: the group Sym({1,2,3}) is abelian iff for all a, b in Sym({123}), a o b = b o a

same, when you conclude from giving a counter-example, do not forget the "there exists"
Thus, there exists a,b in Sym({1,2,3}) such that aob ≠ boa

or ofc using $\exists, \forall$