Question

check part 1 and help part 2, please

Answer 1

give me a hand, please

Answer 2

let me get to a real computer and we can do this
can't really see it on a laptop

Answer 3

ok we have \[\{1,a, b, ab, ba, aba\}\]
with the rules
\[a^2=b^2=1\] and \(aba=bab\)

Answer 4

@satellite73 after u r done helping Hoa, can you please look at my problem: http://openstudy.com/study#/updates/5173404de4b04b62aaef54a0

Answer 5

yes

Answer 6

are you confident about your answers to the first part, or should we check them?

Answer 7

surely I need you check. not confident at all

Answer 8

okay i have to write them on paper hold on one sec

Answer 9

first one looks good
\[a^7b^3a^5b^7=abab\]since \(a^2=b^2=1\) and then since \(bab=aba\) you have \(aaba\) giving \(ba\)

Answer 10

hi friend, what it means? mine is ba and yours is abab

Answer 11

but aba = bab so i replace to get babb = ba, is it right?

Answer 12

yes, your answer is right

Answer 13

that is, \(a(bab)=a(aba)=ba\) which will make the next one easy

Answer 14

since you can group the next one as
\[(abab)(abab)(ab)=(ba)(ba)(ab)=babaab=babb=ba\]

Answer 15

so, mine is ok, too?

Answer 16

yes both 1 and 2 are correct
if you really want to make your life easy, make a 6 by 6 multiplication table for this

Answer 17

that will also help with part 2 and will also make finding the inverses easy

Answer 18

got what you mean. ok part 2

Answer 19

no, sorry, inverse check please

Answer 20

okay
you have the inverses of \(1, a, b\) they are there own inverses
now how about \(ab\)? it is always the case that \((ab)^{-1}=b^{-1}a^{-1}\) not matter what group you are in

Answer 21

since \(b^{-1}a^{-1}=ba\) as \(b\) and \(a\) are their own inverses, you have \((ab)^{-1}=ba\) and so necessarily \((ba)^{-1}=ab\) as inverses are unique

Answer 22

the check is easy enough
\[abba=aa=1\] and \[baab=bb=1\]

Answer 23

satellite73, i lost you

Answer 24

oh sorry lets go slow

Answer 25

you only have 6 elements in this group, so no matter what combination of \(a\) and \(b\) they throw at you, it must be one of the six
\[\{1, a, b, ab, ba, aba\}\] it can't be anything else

Answer 26

yeap

Answer 27

so one way to find the inverse of anything given, is to find the inverses of the six members of the group. then the inverse of whatever is given can be found more easily by figuring out which of those six elements it is. then you will know the inverse

Answer 28

in other words, the best and most systematic way to proceed is to find the inverse of each of the six elements of your group

Answer 29

no, not sure, you guide me, please

Answer 30

you already know the inverse of 3 of the six elements

Answer 31

\[1^{-1}=1\]\[a^{-1}=a\]\[b^{-1}=b\] that much is clear, yes or no?

Answer 32

yeap and (ab)^- = ba as you said above, right?

Answer 33

yes, and as i said above, no matter what group you are in, matrices, functions, integers, abelian, not abelian, whatever, it is always the case that
\[(ab)^{-1}=b^{-1}a^{-1}\] becase
\[abb^{-1}a^{-1}=a1a^{-1}=aa^{-1}=1\]

Answer 34

(ba)^- = a^-b^- = ab too, right?

Answer 35

yes, for exactly the same reason
also, because inverses are unique, so if you know \((ab)^{-1}=ba\) then it is necessarily true that \((ba)^{-1}=ab\)

Answer 36

so the group of inverse is (1, a,b, ba, ab , aba) itself

Answer 37

ok back up a second

Answer 38

the inverse of an element of a group must also be an element of the group. any group is closed. and of course closed under inverses
so it must be the case that the inverse of every element is some other element
at the moment, we are just trying to identify those inverses

Answer 39

so far we have 5 out of the six
we know
\[1^{-1}=1\]\[a^{-1}=a\]\[b^{-1}=b\]\[(ab)^{-1}=ba\]\[(ba)^{-1}=ab\] all that is left is to find \[(aba)^{-1}\]

Answer 40

(aba)^- = (ba)^- a^- = aba

Answer 41

yes, it is its own invere

Answer 42

:)

Answer 43

check is \[(aba)(aba)=ab1ba=abba=a1a=aa=1\]

Answer 44

now to find the inverse of each of those thing written in problem 2, figure out which ones they actually are and you will know the inverse of each

Answer 45

for example, since \(bab=aba\) you know the inverse of \(bab\) is \(aba\)

Answer 46

yes, sir

Answer 47

and since
\[a^3b^4a^7b^{15}=aa^7b^{15}=a^8b^{15}=b\] you know the inverse is \(b\)

Answer 48

I have a question, can i reduce the form first, and then find inverse of irreducible form?

Answer 49

that is exactly what i was trying to say, yes

Answer 50

figure out which of the six it is, then you know the inverse right away

Answer 51

ty let it there, part C please, the hardest one

Answer 52

okay first off the answer
there i are exactly 2 groups of order 6 ( up to isomorphism)
they are \(\mathbb{Z}_6\) and \(S_3\)
\(S_3\) is the smallest non abelian group

Answer 53

yes,

Answer 54

so this is definitely \(S_3\) since this one is not abelian
but your job is to exhibit the isomorphism directly
that is, say what each element gets mapped to

Answer 55

have to make the Carley table?

Answer 56

okay hold on i got the notation mixed up
they are calling this group \(S_3\) and are asking you to show it is isomorphic to \(D_3\)

Answer 57

you can make a multiplication table and exhibit it that way. i guess that is what you are calling a "carley table" but to be honest i have never heard that term
or you can simply say what each element gets mapped to

Answer 58

the elements of this group are
\[S_3=\{1, a, b, b, ab, aba\}\] and the elements of the dihedral group are being called
\[D_3=\{1, r,r^2,f,fr, fr^2\}\] and your job is to say what the mapping is
a huge hint is that each element must get mapped to an element of the same order

Answer 59

i am afraid that won't work

Answer 60

oh, I just know that, what is wrong

Answer 61

lets check for example
\[\phi(a)=r\] and see why it doesn't work

Answer 62

or switch b correspond to f

Answer 63

\[\phi(aa)=\phi(1)=1\] but \(\phi(a)\phi(a)=rr=r^2\neq 1\) so you do not have a homomorphism, because you need \(\phi(a)\phi(a)=\phi(aa)\)

Answer 64

make sure to map each element of \(S_3\) so an element of \(D_3\) with the same order, otherwise it won't work

Answer 65

got what you mean.

Answer 66

you might start by mapping \(a\to f\) since they both have order 2

Answer 67

how to say, it cannot be done in short time , friend. sorry , i am tooo slow

Answer 68

start by saying \(\phi(1)=1, \phi(a)=f\) and then see what you get from that
make sure to also send \(b\) to an element of order 2

Answer 69

b with r

Answer 70

ab = fr

Answer 71

is it ok, first, don't let me go tooo far and then tell me that it doesn't work

Answer 72

no you cannot have \(b\to r\) for the same reason you cannot have \(a\to r\)
both \(a\) and \(b\) have order 2 and \(r\) has order 3

Answer 73

got it , b = r^2

Answer 74

hmm i think \(r^2r^2=r\) so that won't work

Answer 75

ha!!!!

Answer 76

you need to find an element of \(D_3\) with order 2

Answer 77

i would just say it, but i am not familiar with the representation of \(D_3\) as
\[D_3=\{1, r, r^2, f, fr, fr^2\}\] i usually think of it as\[\{1, r, r^2, f_1, f_2, f_3\}\] where \(r\) is a rotation and \(f\) is a flip

Answer 78

try \(ab\to r\) since both \(ab\) and \(r\) have order 3

Answer 79

then you should be able to get the rest from those two

Answer 80

can you draw it down? It is easier when take a look at the graph

Answer 81

try \(\phi(a)=f, \phi(b)=fr, \phi(ab)=r\) and see what you get

Answer 82

we don't have any rule for D2 how can you make it as it is?

Answer 83

you need to know the rules for \(D_3\) that is, the multiplication table

Answer 84

teach me, please, shame on me, you say something about order 2,3.... I fell like foreign language and wonder why my prof never tell me about that

Answer 85

oh maybe it is just the language it am using. let me be more specific

Answer 86

\(a\) has "order 2" means \(a^2=1\)

Answer 87

likewise \(b\) has order 2 because \(b^2=1\)

Answer 88

now \(ab\) has order 3, because \((ab)^3=1\) we know this because
\[(ab)(ab)(ab)=ababab=a(bab)ab=a(aba)ab=aabaab=1b1b=bb=1\]

Answer 89

the "order" of an element \(x\) is the least integer \(n\) so that \(x^n=1\)

Answer 90

amazing, never know

Answer 91

so what i am saying is this: if you want an isomorphism between the two groups, you must map elements of the first group to elements of the second group with the same order

Answer 92

that is why it didn't work to say \(\phi(b)=r\)
because \(\phi(bb)=\phi(1)=1\) but \(\phi(b)\phi(b)=rr=r^2\neq 1\)

Answer 93

but to yours when calculate ba i got frf which is not in D2

Answer 94

since in \(S_3\) you know \(a^2=1\) you have to map it to some element \(x\) of \(D_3\) with \(x^2=1\)

Answer 95

I got what you mean

Answer 96

wait a second
it must be the case that \(frf\in D_3\) since it is closed.
maybe you don't know what it is, but it must be in there