Question

group theory:
f:G->H is a homomorphism
if n is in Z s.t. (n,|H|)=1
prove: if a^n in the kernel of f, then a is in the kernel of f

Answer 1

thought I would tag you two @joemath314159 @KingGeorge
because you two love some group theory:)

Answer 2

You want to work through this problem. I'm not an algebraist, but I have an idea.

Answer 3

@saifoo.khan and @jim_thompson5910 and @satellite73 might be of good good help :P

Answer 4

@zzr0ck3r

Answer 5

sure would love to

Answer 6

what is \[f(a^n)\]

Answer 7

e

Answer 8

and f(a)^n

Answer 9

right

Answer 10

you know what this is <f(a)>

Answer 11

real fast, we proved if f is homomorphism then f(a^-1) = f(a) ^ -1 how can f have an inverse if we do not know its one to one and onto?

Answer 12

yes I do

Answer 13

sorry save the second question till after.

Answer 14

ok

Answer 15

<f(a)> would be the cyclic group generated by f(a)

Answer 16

yes

Answer 17

you know Lagrange's theorem

Answer 18

yes any order of a subgroup divides the order of the parent group

Answer 19

so that tells us that ...

Answer 20

so n divides order of H?

Answer 21

but (n,|H|) = 1

Answer 22

well |<f(a)>| divides |H|

Answer 23

right

Answer 24

need to show |<f(a)>| divides n...then we are done

Answer 25

why ?

Answer 26

why it divides or why we are done?

Answer 27

why we are done

Answer 28

(n,H)=1

and n and |h| will have a common factor...thus |<f(a)>|=1 and hence f(a)=e

Answer 29

|H|

Answer 30

ahhhh very clever

Answer 31

if a|b and a|c and (b,c)=1 then a=1

Answer 32

right right

Answer 33

so we are using the fact that f(a)^n = e and thus f(a) has finite order and thus forms a cyclic subgroup

Answer 34

you should have a theorem or corollary that states that if

\(a^k=e\) then \(|a|\) divides \(k\)

Answer 35

yes for sure

Answer 36

that is all you need

Answer 37

wow tyvm. tnx for dragging me along also. it helps allot.

Answer 38

np

Answer 39

what about this

"real fast, we proved if f is homomorphism then f(a^-1) = f(a) ^ -1 how can f have an inverse if we do not know its one to one and onto?"

Answer 40

yes

Answer 41

very confused by that

Answer 42

this is still a group homomorphism?

Answer 43

yes

Answer 44

so if \(f:G\to H\) then f(a) is an element of the group H. All group elements have an inverse

Answer 45

ahh so the inverse is an inverse of the element f(a) in H so f(a)^-1 is in H not G?
talking about f:g->h

Answer 46

call if \(f^{-1}(a)\)

Answer 47

so its a notation thing that is confusing me

Answer 48

ahh so f^-1(x) is a fuction and f(x)^-1 is an element? in this context?

Answer 49

well I guess that question is redundant, but I'm pretty sure I understand

Answer 50

you know the proof

Answer 51

maybe that would help you understand if you saw that

Answer 52

yeah, I do. I was just thinking that. I think someone in class confused me today because they said f had an inverse. I will clear it up with them and take all the credit:)
jk

Answer 53

wow for not being an algebraist you have quite a memory of what the proofs look like...

Answer 54

did you want me to show you or do you know it...it is short

Answer 55

I do know it, ty.

Answer 56

ok

Answer 57

I took quite a few algebra class in graduate school, but most of my course work was in probability and mathematical statistics

Answer 58

you an actuary?

Answer 59

I'm a math professor

Answer 60

ahhh it all makes sense now

Answer 61

my Ph.D. is in mathematical statistics

Answer 62

very cool, I want to be you - the stats:) + something else in exchange

Answer 63

but i really love algebra...beautiful subject

Answer 64

yeah I really like it so far.
this is just 300 level group theory but I have to do 2 term 400 sequence in the fall, with 400 real analysis as well.

Answer 65

nice...real analysis is fun too

Answer 66

I took the 300 level one, for us its 311 and 312. I like it, but I like the algebra much more. I am ok with a summer of no deltas or epsilons.

Answer 67

I'm not a huge fan of analysis either to be honest. I much prefer algebra.

Answer 68

lol

Answer 69

I find algebra to be more elegant

Answer 70

for sure. there is just something sort of hand-wavey about calculus. (not really but ...)

Answer 71

I understand what you are saying

Answer 72

Same.

Answer 73

@Zarkon how do we know <f(x)> forms a cyclic subgroup?

Answer 74

by definition <a> is the cyclic group generated by a

Answer 75

so If group G has finite order than any element in G creates a cyclic subgroup?

Answer 76

THeorem:

Let \(G\) be a group, and let \(a\) be any element of \(G\). Then, \(<a>\) is a subgroup of \(G\)

Answer 77

great.. ty

Answer 78

np