Why for a group homomorphism

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shmoe

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You're trying to prove that a group isomorphic to the integers is cyclic? Are the integers cyclic?
 

quasar987

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Mh, yeah it's that trivial huh?

In the book I'm using, there is a theorem stated before the one I'm trying to prove that says that 'Isomorphisms preserve all algebraic properties'. In particular, if f:G-->G' is an isomorphism, then G' is cyclic iff G is cyclic.

But I couldn't prove that theorem, because what are algebraic properties exactly?
 
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If x is a generator of G, then f(x) is a generator of G'.

The converse follows from the fact that f has an inverse.
 

quasar987

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Yeah, but is there a way to prove the general statement "Isomorphisms preserve all algebraic properties" ?
 

matt grime

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yes, since the algebraic properties of groups are by definition those things preserved by isomorphisms. isomorphic groups are identical as groups in terms of their behaviour as groups.

given a class of objects sharing some common definition an isomorphism is something that preserves exactly those properties that can eb deduced from the definition. that's why we study isomorphisms.

So, an iso of groups is a bijection between G and H such that f(xy)=f(x)f(y) and f(x^-1}=f(x)^{-1}. G and H are idenntical oas groups as the identification x <--> f(x) shows so all group theoretic (ie the algebraic properties of groups) are preserved by definition.
 
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quasar987

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If an algebraic property is by definition a property that is preserved by isomorphisms, isn't proving that all algebraic properties are preserved by isomorphisms futile?

And in identifying the algebraic properties of groups, one would have to test them one by one: Is the order preserved? Is the property of being cyclic preserved. If H is a normal subgroup in G, is f(H) normal in G' too?, etc.
 

matt grime

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An isomorphism is exactly something that preserves the properties of a group that are a direct consequence of it being a group, or drectly are defined by group theoretic terms alone, and you can prove all of these very easily from the definition of an iso which is why they are good questions to set beginning students.

Let f be an iso from G to H

Examples.

if x is conjugate to y f(x) is conjugate to f(y). Proof: there is an z such that yz=zz hence f(y)f(z)=f(z)f(x)

if x has order n so does f(x)
G is abelian if and only if H is abelian

K is normal in G if and only if f(K) is normal in H

G is cyclic iff and only if H is

and so on and so on.

The definition of iso is straight forward, it is up to you it preserves properties of groups. those properties it preserves are exactly those things that we think of as being innately algebraic FOR GROUPS, that is the intrinsic properties of groups
 

quasar987

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Concerning another problem:

If a is an element of a group G, there is always a homomorphism from Z to G which sends 1 to a. When is there a homomorphism from [itex]\mathbb{Z}_n[/itex] to G which sends [1] to a? What are the homomorphisms from [itex]\mathbb{Z}_2[/itex] to [itex]\mathbb{Z}_6[/itex]?
I answered that there is always such an homomorphism in the person of f([m]) = a^m. The homomorphisms from Z_2 to Z_6 are [itex]f_i([m]) = i[m][/itex], [itex]i = 0,1,...,5[/itex]

This looked good to me though I was not entirely sure there exists no other homomorphisms from Z_2 to Z_6. But then the next question is

Suppose G is a group and g is an element of G, g [itex]\neq[/itex] e. Under what conditions on g is there a homomorphism f : Z_7 --> G with
f([1]) = g ?
I would answer 'always', but it wouldn't make sense to ask that question if the answer to the general question 'When is there a homomorphism from [itex]\mathbb{Z}_n[/itex] to G which sends [1] to a?' is 'always'.

What's wrong here?
 

matt grime

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f is a homo. if x has order n the e=f(x^n)=f(x)^n so what is the order of f(x)? when you've figure that out redo that last post
 

quasar987

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You've said it yourself in post #32: n. But how is this related to this problem ?! :confused:
 

matt grime

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no, post 32 was about isomorphisms. think about it. if g is an element of any group and g^n=e what does that tell you about the order of g?

from what you;ve said you appera to think that there is a non-zero homomorphism from Z_7 to Z, and there isn't.
 

quasar987

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The order of g divides n.

How is f([m]) = a^m not an homomorphism from Z_7 to Z?
 

matt grime

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What is a? why does the order m, which divides 7, divide the order of a^m? how can it? there are no elements of finite order in Z apart from 0 are there? so the only homo from Z_7 to Z is the trivial one.
 
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matt grime

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one thing that would really help you is to learn the isomorphism theorems (and don'r sey they are in another chapter. one can prove all these things without tem but to understand why they are true it is ehlpful to know them).

if f is a homo from G to H it means that there is a quotient group of G isomorphically embedded in or has a copy inside or is a subgroup of H. Z_7 has no non-trivial quotients os it's clear that theer can be no isomorphic embedding of Z_7 in Z, or less fancilly, Z_7 is not a subgroup of Z, though you have stated it is.
 

quasar987

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a is some element in Z.

Defined in this way, f([m]+[l]) = f([m]) f([l]), so f is an homomorphism. That's how I see it. How do you?
 
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But is it even a function if you don't place more restrictions on a? [8] = [1] (in Z_7), so f([8]) = f([1]), i.e a^8 = a (if f were a function)...
 

matt grime

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oh come on! i've already explained it isn't a homomorphism unless a=0 at which point it is trivial. in the sepciifc example you gave you were asked to send 1 in Z_7 to g in G where g=/=e so that rules out the trivial case.


Let is use addition since Z is an additive group, as is Z_7

you are sending [m] in Z_7 to am in Z

this isn't a homomorphism, and you should prove it. HINT 3+4=0 in Z_7 and doesn't in Z
 
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quasar987

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matt, let's start with the first problem, which is

If a is an element of a group G, there is always a homomorphism from Z to G which sends 1 to a. When is there a homomorphism from Z_n to G which sends [1] to a? What are the homomorphisms from Z_2 to Z_6?
Muzza said:
But is it even a function if you don't place more restrictions on a? [8] = [1] (in Z_7), so f([8]) = f([1]), i.e a^8 = a (if f were a function)...
Oh, so the problem is that f is not even a function! Interesting... Well it would be if a is e!

Ok, so now to the first question, I can answer that f, defined as

[itex]f([m])=a^m[/itex]

(supposing G is multiplicative) is an homomorphism that sends [1] to a always if a=e and under the condition that n divides o(G) if a [itex]\neq[/itex] e. That's that for this particular function. Now, are there other possible functions that are homomorphisms and that send [1] to a? I don't know.

To the other, I would say that the homomorphisms from Z_2 to Z_6 are

[tex]\{f_i([m]) = 3i[m]\}_{i \in \mathbb{Z}}[/tex]
 

matt grime

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you appear to have an infintely family of homomorphisms, are you claiming that they are different?

Suppose the map from Z_2 to Z is an isomorphism, then it must send 0 to 0 and 1 to an element of order 2, and there is only one of those, 3. If it isn't an isomorphism then it must send 0 and 1 to 0 so there are two homos.

try it if we think of maps Z_3 to Z_6
 

quasar987

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No, I don't claim that; the f_i's for odd i's all do the same thing, namely send [m] to [3] and the f_i's for even i's all do the same thing, namely send [m] to [0].

Let me now ponder on your isomorphism argument there.
 

quasar987

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I can't crack the code :tongue:

matt grime said:
Suppose the map from Z_2 to Z is an isomorphism
The map? What map? Also, is it not impossible to have an isomorphism btw a group of order 2 and one of infinite order?

matt grime said:
, then it must send 0 to 0 and 1 to an element of order 2, and there is only one of those, 3.
Please use the [] notation for elements of Z_n in order to avoid confusion, thx.

matt grime said:
If it isn't an isomorphism then it must send 0 and 1 to 0 so there are two homos.
Two homos! Could you extrapolate the logical steps leading to that conclusion?
 

matt grime

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apologies, i meant to say "suppose f is an injective homomorphism, ie its kernel is {e}, from Z_2 to Z_6" so there were at least 2 errors there.

no, i won;t use brackets for Z_n since it is unnecessary.
 

quasar987

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quasar987 said:
Ok, so now to the first question, I can answer that f, defined as

[itex]f([m])=a^m[/itex]

(supposing G is multiplicative) is an homomorphism that sends [1] to a always if a=e and under the condition that n divides o(G) if a [itex]\neq[/itex] e. That's that for this particular function. Now, are there other possible functions that are homomorphisms and that send [1] to a? I don't know.
How do we show that all homomorphisms that send [1] to a are of the form above?

edit: Or rather, how do we show that f above is the only homo that sends [1] to a.
 
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AKG

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[itex]\mathbb{Z}_n[/itex] is cyclic, so to find what a homomorphism does to any element m in Zn, you only need to know what it does to 1, since 1 generates the group:

f(m) = f(1m)
= (f(1))m since f is a homomorphism
= am since we stipulated that f sends 1 to a

Note that normally, ab means that we multiply a, b times. However, when dealing with groups, it is common to refer to the group operation as multiplication (regardless of what it is - addition, addition mod n, multiplication, composition, etc.) and it is common to use "multiplicative notation," regardless of what the operation is. So 1m means that we do the group operation to 1, m times. The group operation happens to be addition modulo n, so we might prefer to write m(1) instead of 1m, but as long as you keep in mind that there is only one operation that concerns us, then the notation 1m meaning to add 1 to itself m times shouldn't confuse you.

Anyways, now you should be able to see that f must send m to am. There is only really one restriction that needs to be placed in order for f to be a function. Once this restriction is in place, not only will f be a function, but by the way we defined it, it will automatically be a homomorphism. Can you figure out this restriction? n need not divide the order of G. In fact, G could be an infinite group, in which case it may not even make sense to say whether n divides |G| or not.
 

quasar987

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It must be that the order of a divides n: o(a)|n
 
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