Question

Group Theory Q. if G1 is a cyclic group with generator a, prove that G2 is also a cyclic group, with generator f(a).

I know I need to show f(ab)=f(a)f(b). but im not sure how to do so given that its general....a cyclic group could be modulo, group of symmetries, etc. Where do I start?

Answer 1

there must be something missing from this problem

Answer 2

what is \(G_2\) in relation to \(G_1\)

Answer 3

oh...yes

let f: G1->G2 be an isomorphism

Answer 4

sorry about that omission

Answer 5

oh that is rather a lot of information
it means they are the same group, only the elements are called different things

Answer 6

i see. so I just need to show an element from G1 and an element from G2?

Answer 7

write down exactly what " \(a\) generates the cyclic group \(G_1\)" means

Answer 8

in G1, <a> = {a1,a2,...an} so f(an)=bn, where bn is some element in g2?? i'm not confindent in the notation

Answer 9

\(a\) generates the cyclic group \(G_1\) means any element of \(G_1\) is \(a^k\) for some \(k\) i.e. if \(g\in G_1\) then there is a \(k\) such that \(g=a^k\)

Answer 10

and the notation is \(G_1=<a>\) yes

Answer 11

oooohhh. yaaa. i forgot that notation for cyclic groups. thank you.

Answer 12

now what exactly do you want to show?

Answer 13

if G1 is a cyclic group with generator a, i need to prove that G2 is also a cyclic group, with generator f(a)

Answer 14

i mean "exactly"
that is what most of the work for the beginning group theory problems amount to
write down precisely what you want to show, then it will probably be obvious how to get it

Answer 15

i think it got it. thank you so much. its so easy to not know where to start with these problems, as it is my first upper level math course.

Answer 16

maybe it is not clear what i mean
forget about how to show it for a second
what would " G2 is also a cyclic group, with generator f(a)" mean precisely?

Answer 17

that there is an element in G2, say a, s.t. <a>={a1, a2,...an} so i need to show if y is an element of G2, then y=f(a)=b^k=a^-1??

Answer 18

it is the same as what i wrote above for \(G_1\) but replacing \(a\) with \(f(a)\)

Answer 19

i.e. if \(g\in G_2\) then there is an integer \(k\) with \(g=(f(a))^k\)

Answer 20

that will make \(f(a)\) a generator of \(G_2\)

Answer 21

its just simply substituting a for f(a)?

Answer 22

f(a) for a?

Answer 23

right, that is what you need to show

Answer 24

so lets pick an arbitrary element \(g\in G_2\)

Answer 25

there must be some element \(g'\in G_1\) such that \(f(g')=g\)
why?

Answer 26

because its bijective?

Answer 27

right

Answer 28

well onto is enough, but yes, bijective is good too
now since \(G_1\) is cyclic, what do we know about that \(g'\in G_1\) ?

Answer 29

i guess the bijective part allows you to say there is a unique element \(g'\in G_1\) such that \(f(g')=g\)

Answer 30

in any case, what must be true about \(g'\) ?

Answer 31

this is where you use the fact that \(G_1\) is cyclic and generated by \(a\)

Answer 32

for g' in G1, there is f(g')=a

Answer 33

no

Answer 34

lets go slow
\(g'\in G_1\) right? and \(G_1\) is generated by \(a\)
this means that \(g'=a^k\) for some \(k\) right?

Answer 35

ok

Answer 36

btw \(a\in G_1\) not in \(G_2\)
it is \(f(a)\in G_2\)

Answer 37

i just drew a diagram to keep track

Answer 38

ok so since \(g'=a^k\) what is another way to write \(f(g')=f(a^k)\) using the fact that \(f\) is a homomorphism?

Answer 39

i haven't learned homomorphism yet, but i suspect f(a^k)=f(a^k)?

Answer 40

you must have learned about homomorphisms, because an isomorphism is a homomorphism
\(f:G_1\to G_2\) is a homomorphism means for all \(x, y \in G_1\) you have
\[f(x\circ y)=f(x)\times f(y)\] where \(\circ\) is the multiplication in \(G_1\) and \(\times \) is the multiplication in \(G_2\)
sometimes this is abbreviated as \(f(xy)=f(x)f(y)\) but the two multiplications can be different

Answer 41

homomorphisms is chapter 14. we are on chapter 9 now...isomorphisms.

Answer 42

this tells you that
\[f(a^k)=\left(f(a)\right)^k\]

Answer 43

ok well that is one property of a isomorphism as well
a homomorphism need not be a bijection

Answer 44

is it clear what the difference between the left and right had side of \[f(a^k)=\left(f(a)\right)^k\] is?

Answer 45

hi @joojoo :)

Answer 46

yes. the left side is cyclic in G1 and right side is cyclic in G2

Answer 47

hi zz :) group theory is painful.

Answer 48

not exactly
this \(f(a^k)\) means take \(a\) raise it to the power of \(k\) and then send it via \(f\) to \(G_2\)

Answer 49

yes, i understand that

Answer 50

this \(\left(f(a)\right)^k\) means take \(a\), map it via \(f\) to \(G_2\) and raise the result to the power of \(k\)

Answer 51

which means essentially we are done

Answer 52

i guess that is what i was trying to say...a^k comes from G1 and the left is what lands in G2

Answer 53

ok good

Answer 54

i mean the right

Answer 55

so lets recap and put it all in one package

Answer 56

thank you, satellite...i suspect i will have to stare at this for a while until it sinks in...its not difficult...but i just trip up on writing it all correctly and keeping my elements straight

Answer 57

@joojoo told you this place rules:)

Answer 58

\(G_1\) is cyclic with generator \(a\) and there is an isomorphism \(f:G_1\to G_2\)
you want to prove that \(G_2\) is also cyclic with generator \(f(a)\)

what this means specifically is that given any \(g_2\in G_2\) you can write \(g_2=\left (f(a)\right)^k\) for some \(k\)

Answer 59

yes, yes it does, zz!! isomorphs are not fun right now.

Answer 60

since \(f\) is an isomorphism, there is a \(g_1\in G_1\) such that \(f(g_1)=g_2\)

Answer 61

since \(G_1\) is cyclic that means \(g_1=a^k\) for some \(k\)

Answer 62

and since \(f\) is an isomorphism, \(f(g_1)=f(a^k)=\left(f(a)\right)^k=g_2\) which is exactly what you needed to show

Answer 63

satellite, got it. i will need to practice writing it out on my own....i suspect several times.

Answer 64

ah ha. i get it.

Answer 65

when you do any of these types of problems, the first thing you need to do is write down the definitions exactly, i.e. figure out exactly what each statement means without the verbiage "isomorphism", "cyclic" etc.

then write down specifically what you know and specifically what you need to show

if you are confused, that is fine, that is why you are given the problems. the "prove" part at the beginning should then be more or less obvious

Answer 66

thank you for your advice. i spend a good amount of time on problems, and more often, i feel like im dragging thru mud. i guess practice will help.

Answer 67

yes, it will help and later you will look at a problem like this one and think "oh nothing to it, it is much easier than finding the derivative of \(e^x\sin(x)\)

Answer 68

i hope so :) i really appreciate your patience and help!!

Answer 69

yw